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A particle is dropped from a height \(H\). The de Broglie wavelength of the particle as a function of height is proportional to
(d) De Broglie wavelength: \(\lambda=\frac{{h}}{{p}}\), where \({h}\) is Planck’s constant and \(p\) is momentum.
Kinetic energy: \(K E=\frac{1}{2} m v^2=\frac{p^2}{2 m}\), where \(m\) is mass and \(v\) is velocity.
Potential energy at height \({H}: P \mathbf{P E}={m g} {H}\), where \({g}\) is acceleration due to gravity.
Conservation of energy: \(P E=K E\) when the particle reaches the ground.
Equate potential energy at height \({H}\) to kinetic energy just before impact:
\(m g H=\frac{1}{2} m v^2\)
\(
v=\sqrt{2 g H}
\)
Express momentum \(p\) in terms of \(m\) and \(v\) :
\(p=m v=m \sqrt{2 g H}\)
Substitute \(p\) into the de Broglie wavelength formula:
\(
\lambda=\frac{h}{m \sqrt{2 \mathrm{gH}}}
\)
Identify the proportionality between \({\lambda}\) and \({H}\) :
\(\lambda \propto \frac{1}{\sqrt{H}}=H^{-\frac{1}{2}}\)
The de Broglie wavelength of the particle is proportional to \({H}^{-\frac{1}{2}}\).
The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
(b) Planck’s constant: \(h=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\)
Speed of light: \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Conversion factor: \(1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}\)
Calculate the wavelength of the photon using the given energy and the formula \(E=h \frac{c}{\lambda}\).
Convert the energy from MeV to Joules
\(
\begin{aligned}
& E=1 \mathrm{MeV}=1 \times 10^6 \mathrm{eV} \\
& E=1 \times 10^6 \mathrm{eV} \times 1.602 \times 10^{-19} \frac{\mathrm{~J}}{\mathrm{eV}} \\
& E=1.602 \times 10^{-13} \mathrm{~J}
\end{aligned}
\)
Calculate the wavelength using the formula \(E=h \frac{c}{\lambda}\)
\(
\begin{aligned}
& \lambda=\frac{h c}{E} \\
& \lambda=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{1.602 \times 10^{-13} \mathrm{~J}} \\
& \lambda=\frac{19.878 \times 10^{-26}}{1.602 \times 10^{-13} \mathrm{~m}} \\
& \lambda=12.408 \times 10^{-13} \mathrm{~m} \\
& \lambda=1.2408 \times 10^{-12} \mathrm{~m}
\end{aligned}
\)
Convert the wavelength from meters to nanometers
\(
\begin{aligned}
& \lambda=1.2408 \times 10^{-12} \mathrm{~m} \times \frac{10^9 \mathrm{~nm}}{1 \mathrm{~m}} \\
& \lambda=1.2408 \times 10^{-3} \mathrm{~nm} \\
& \lambda \approx 1.2 \times 10^{-3} \mathrm{~nm}
\end{aligned}
\)
The wavelength of the photon is approximately \(1.2 \times 10^{-3} \mathrm{~nm}\).
Consider a beam of electrons (each electron with energy \(E_0\) ) incident on a metal surface kept in an evacuated chamber. Then
(d) If a beam of electrons of having energy \(E_0\) is incident on a metal surface kept in an evacuated chamber.
The electrons can be emitted with maximum energy \({E}_0\) (due to elastic collision) and With any energy less than \(E_0\), when part of incident energy of electron is used in liberating the electrons from the surface of metal.
Consider the figure given below. Suppose the voltage applied to \(A\) is increased. The diffracted beam will have the maximum at a value of \(\theta\) that
Â
(c) The de-Broglie wavelength associated with electron is
\(
\lambda=\frac{12.27}{\sqrt{V}} Ã… \dots(i)
\)
where \(V\) is the applied voltage.
Using the Bragg’s formula we can determine the wavelength of these waves. If there is a maxima of the, diffracted electrons at an angle \(\theta\), then
\(
2 d \sin \theta=\lambda \dots(ii)
\)
From Eq. (i), we note that if \(V\) is inversely proportional to the wavelength \(\lambda\). i.e., \(V\) will increase with the decrease in \(\lambda\).
From Eq. (ii), we note that wavelength \(\lambda\) is directly proportional to \(\sin \theta\) and hence \(\theta\).
So, with the decrease in \(\lambda, \theta\) will also decrease.
Thus, when the voltage applied to \(A\) is increased. The diffracted beam will have the maximum at a value of \(\theta\) that will be less than the earlier value.
A proton, a neutron, an electron and an \(\alpha\)-particle have the same energy. Then their de Broglie wavelengths compare as
(b) De-Broglie wavelength: According to de-Broglie theory, the wavelength of de-Broglie wave is given by
\(
\lambda=\frac{h}{p}=\frac{h}{m v}=\frac{h}{\sqrt{2 m E}} \Rightarrow \lambda \propto \frac{1}{p} \propto \frac{1}{v} \propto \frac{1}{\sqrt{E}}
\)
Where \(h=\) Plank’s constant, \(m=\) Mass of the particle, \(v=\) Speed of the particle, \(E=\) Energy of the particle.
The smallest wavelength whose measurement is possible is that of \(\gamma\)-rays.
The wavelength of matter waves associated with the microscopic particles like electron, proton, neutron, \(\alpha\)-particle etc. is of the order of \(10^{-10} \mathrm{~m}\)
We conclude from above that the relation between de-Broglie wavelength \(\lambda\) and kinetic energy \(K\) of the particle is given by
\(
\lambda=\frac{h}{\sqrt{2 m K}}
\)
Here, for the given value of energy \(K, \frac{h}{\sqrt{2 K}}\) is a constant.
Thus, \(\quad \lambda \propto \frac{1}{\sqrt{m}}\)
\(
\therefore \quad \lambda_p: \lambda_n: \lambda_e: \lambda_\alpha=\frac{1}{\sqrt{m_p}}: \frac{1}{\sqrt{m_n}}: \frac{1}{\sqrt{m_e}}: \frac{1}{\sqrt{m_\alpha}}
\)
Comparing wavelength of proton and neutron, \(m_p=m_n\), hence \(\lambda_p=\lambda_n\)
As, \(\quad m_\alpha>m_p\) therefore \(\lambda_\alpha<\lambda_p\)
As, \(\quad m_e>m_n\) therefore \(\lambda_e<\lambda_n\).
Hence, \(\quad \lambda_\alpha<\lambda_p=\lambda_n<\lambda_e\).
An electron is moving with an initial velocity \(\mathbf{v}=v_0 \hat{\mathbf{i}}\) and is in a magnetic field \(\mathbf{B}=B_0 \hat{\mathbf{j}}\). Then it’s de Broglie wavelength
(a) According to the problem, \(\vec{v}=v_0 i\) and \(\vec{B}=B_0 j\) Magnetic force on moving electron \(=-e\left[v_0 i \times B_0 j\right] \Rightarrow=-e v_0 B_0 k\)
As this force is perpendicular to \(\vec{v}\) and \(\vec{B}\), so the magnitude of \(v\) will not change, i.e. momentum ( \(p=m v\) ) will remain constant in magnitude. Hence, de-Broglie wavelength \(\lambda=\frac{h}{m v}\) remains constant.
An electron (mass \(m\) ) with an initial velocity \(\mathbf{v}=v_0 \hat{\mathbf{i}}\left(v_0>0\right.\) ) is in an electric field \(\mathbf{E}=-E_0 \hat{\mathbf{i}}\left(E_0=\right.\) constant \(\left.>0\right)\). It’s de Broglie wavelength at time \(t\) is given by
(a)
The force on an electron in an electric field is given by \(\vec{F}=-e \vec{E}\), where \(-{e}\) is the charge of the electron.
Newton’s second law: \(\overrightarrow{{F}}={m} \vec{a}\)
The de-Broglie wavelength is given by \(\lambda=\frac{{h}}{{p}}=\frac{{h}}{{m} {v}}\), where \({h}\) is Planck’s constant, \(p\) is momentum, and \(v\) is velocity.
Find the acceleration of the electron
The force on the electron is \(\vec{F}=-e \vec{E}=-e E_0 \hat{i}\).
Using Newton’s second law, \(\vec{F}=m \vec{a}\), so \(m \vec{a}=-e E_0 \hat{i}\).
Therefore, the acceleration is \(\vec{a}=\frac{-e E_0}{m} \hat{i}\).
Find the velocity of the electron at time \(t\)
Using the equation of motion, \(\vec{v}=\vec{v}_0+\vec{a} t\).
Substituting the given values, \(\vec{v}=v_0 \hat{i}+\frac{-c E_0}{m} \hat{i t}\).
Thus, \(\vec{v}=\left(v_0-\frac{e E_0 t}{m}\right) \hat{i}\).
Calculate the de-Broglie wavelength at time \(t\)
The de-Broglie wavelength is given by \(\lambda=\frac{h}{m v}\).
Substituting the velocity at time \(t, \lambda=\frac{h}{m\left(v_0-\frac{e E_0 t}{m}\right)}\).
Simplifying, \(\lambda=\frac{h}{m v_0-e E_{0^t}}\).
We can also write it as \(\lambda=\frac{h}{m v_0\left(1-\frac{e F_0 t}{m v_0}\right)}\).
The initial de-Broglie wavelength is \(\lambda_0=\frac{h}{m v_0}\).
Therefore, \(\lambda=\frac{\lambda_0}{1-\frac{e F_0 t}{m \nu_0}}\).
The de-Broglie wavelength of the electron at time \(t\) is \(\frac{\lambda_0}{1-\frac{e E_0 t}{m \nu_0}}\).
An electron (mass \(m\) ) with an initial velocity \(\mathbf{v}=v_0 \hat{\mathbf{i}}\) is in an electric field \(\mathbf{E}=E_0 \hat{\mathbf{j}}\). If \(\lambda_0=h / m v_0\), it’s de Broglie wavelength at the time \(t\) is given by
(c) According to the problem de-Broglie wavelength of electron at time \(\mathrm{t}=0\).
is \(\lambda_0=\frac{h}{m v_0}\).
Electrostatic force on electron in electric field is
\(
\vec{F}_e=-e \vec{E}=-e E_0 \hat{j}
\)
The acceleration of electron, \(\vec{a}=\frac{\vec{F}}{m}=-\frac{e E_0}{m} \hat{\mathrm{j}}\)
It is acting along negative \(y\)-axis.
The initial velocity of electron along \(x\)-axis, \(v_{x_0}=v_0 \hat{i}\).
This component of velocity will remain constant as there is no force on electron in this direction.
Now considering \(y\)-direction. Initial velocity of electron along \(y\)-axis, \(v_{y_0}=0\).
Velocity of electron after time \(t\) along \(y\)-axis,
\(
v_y=0+\left(-\frac{e E_0}{m} \hat{\mathrm{j}}\right) t=-\frac{e E_0}{m} t \hat{\mathrm{j}}
\)
Magnitude of velocity of electron after time \(t\) is
\(
\begin{aligned}
v & =\sqrt{v_x^2+v_y^2}=\sqrt{v_0^2+\left(\frac{-e E_0}{m} t\right)^2} \\
\Rightarrow \quad & =v_0 \sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}
\end{aligned}
\)
de-Broglie wavelength, \(\lambda^{\prime}=\frac{h}{m v}\)
\(
\begin{aligned}
& =\frac{h}{m v_0 \sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}=\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}} \\
\Rightarrow \lambda^{\prime} & =\frac{\lambda_0}{\left(1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}\right)}
\end{aligned}
\)
Relativistic corrections become necessary when the expression for the kinetic energy \(\frac{1}{2} m v^2\), becomes comparable with \(m c^2\), where \(m\) is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?
(c, d) The wavelength of de-Broglie wave is given by
\(
\begin{aligned}
& \lambda={h} / {p}={h} / {mv} \\
& \text { Here, } {h}=6.6 \times 10^{-34} \mathrm{Js} \\
& \text { and for electron, } {m}=9 \times 10^{-31} \mathrm{~kg}
\end{aligned}
\)
In option (a), \(\lambda_1=10 \mathrm{~nm}=10 \times 10^{-9} \mathrm{~m}=10^{-8} \mathrm{~m}\)
\(
\begin{aligned}
\Rightarrow \quad v_1 & =\frac{6.6 \times 10^{-34}}{\left(9 \times 10^{-31}\right) \times 10^{-x}} \\
& =\frac{2.2}{3} \times 10^5 \approx 10^5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
In option (b), \(\lambda_2=10^{-1} \mathrm{~nm}=10^{-1} \times 10^{-9} \mathrm{~m}=10^{-10} \mathrm{~m}\)
\(
\Rightarrow \quad v_2=\frac{6.6 \times 10^{-34}}{\left(9 \times 10^{-31}\right) \times 10^{-10}} \approx 10^7 \mathrm{~m} / \mathrm{s}
\)
In option (c). \(\lambda_3=10^{-4} \mathrm{~nm}=10^{-4} \times 10^{-9} \mathrm{~m}=10^{-13} \mathrm{~m}\)
\(
\Rightarrow \quad v_3=\frac{6.6 \times 10^{-34}}{\left(9 \times 10^{-31}\right) \times 10^{-13}} \approx 10^{10} \mathrm{~m} / \mathrm{s}
\)
In option (d), \(\lambda_4=10^{-6} \mathrm{~nm}=10^{-6} \times 10^{-9} \mathrm{~m}=10^{-15} \mathrm{~m}\)
\(
\Rightarrow \quad v_4=\frac{6.6 \times 10^{-34}}{\left(9 \times 10^{-31}\right) \times 10^{-15}}=10^{12} \mathrm{~m} / \mathrm{s}
\)
Thus. options (c) and (d) are correct as \(v_3\) and \(v_4\) is greater than \(3 \times 10^8 \mathrm{~m} / \mathrm{s}\).
Two particles \(\mathrm{A}_{1 \mathrm{~s}}\) and \(\mathrm{A}_2\) of masses \(m_1, m_2\left(m_1>m_2\right)\) have the same de Broglie wavelength. Then
(a, c) We know that de-Broglie wavelength \(\lambda=\frac{h}{m v}\) where, \(m v=p\) (momentum) of the particle
\(\Rightarrow\) But we can express wavelength \(\lambda=\frac{h}{p} \Rightarrow p=\frac{h}{\lambda}\) Here. \(h\) is Planck constant.
Hence, \(\quad p \propto \frac{1}{\lambda} \Rightarrow \frac{p_1}{p_2}=\frac{\lambda_2}{\lambda_1}\)
But particles have the same de-Broglie wavelength. \(\left(\lambda_1=\lambda_2\right)=\lambda\).
Then. \(\quad \frac{p_1}{p_2}=\frac{\lambda}{\lambda}=1 \Rightarrow p_1=p_2\)
Thus, their momenta is same.
Also,
\(
\begin{aligned}
E & =\frac{1}{2} m v^2=\frac{1}{2} m v^2 \frac{m}{m} \\
& =\frac{1}{2} \frac{m^2 v^2}{m}=\frac{1}{2} \frac{p^2}{m}
\end{aligned}
\)
As \(p\) is constant, \(E \propto \frac{1}{m}\)
\(
\therefore \quad \frac{E_1}{E_2}=\frac{m_2}{m_1}<1 \Rightarrow E_1<E_2
\)
The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is \(v_e=\frac{c}{100}\). Then
(b, c) The de Broglie wavelength \((\lambda)\) of a particle is given by the formula:
\(
\lambda=\frac{h}{p}
\)
where \(h\) is Planck’s constant and \(p\) is the momentum of the particle.
The momentum of the electron \(\left(p_e\right)\) can be expressed as:
\(
p_e=m_e v_e
\)
where \(m_e\) is the mass of the electron and \(v_e\) is its speed. Given \(v_e=\frac{c}{100}\), we have:
\(
p_e=m_e\left(\frac{c}{100}\right)
\)
Thus, the de Broglie wavelength of the electron \(\left(\lambda_e\right)\) is:
\(
\lambda_e=\frac{h}{p_e}=\frac{h}{m_e\left(\frac{c}{100}\right)}=\frac{100 h}{m_e c}
\)
According to the problem, the de Broglie wavelength of the photon \(\left(\lambda_p\right)\) is twice that of the electron:
\(
\lambda_p=2 \lambda_e=2\left(\frac{100 h}{m_e c}\right)=\frac{200 h}{m_e c}
\)
The energy of the photon \(\left(E_p\right)\) is given by:
\(
E_p=\frac{h c}{\lambda_p}
\)
Substituting for \(\lambda_p\) :
\(
E_p=\frac{h c}{\frac{200 h}{m_e c}}=\frac{m_e c^2}{200}
\)
The kinetic energy of the electron \(\left(E_e\right)\) is given by:
\(
E_e=\frac{1}{2} m_e v_e^2=\frac{1}{2} m_e\left(\frac{c}{100}\right)^2=\frac{m_e c^2}{20000}
\)
Now, we can find the ratio of the energies:
\(
\frac{E_e}{E_p}=\frac{\frac{m_e c^2}{20000}}{\frac{m_e c^2}{200}}=\frac{200}{20000}=\frac{1}{100}
\)
Photons absorbed in matter are converted to heat. A source emitting \(n\) photon/sec of frequency \(\nu\) is used to convert 1 kg of ice at \(0^{\circ} \mathrm{C}\) to water at \(0^{\circ} \mathrm{C}\). Then, the time \(T\) taken for the conversion
(a, b, c) We know that energy spent to con\nuert ice into water
\(
\begin{array}{ll}
& E_{\text {absorb }}=\text { mass } \times \text { latent heat } \\
\therefore \quad & E=m L=(1000 \mathrm{~g}) \times(80 \mathrm{cal} / \mathrm{g}) \\
& E=80000 \mathrm{cal}
\end{array}
\)
Energy of photons used \(=n T \times E=n T \times h \nu [\because E=h \nu]\)
So, \(n T h \nu=m L \Rightarrow T=\frac{m L}{n h \nu}\)
\(\therefore \quad T \propto \frac{1}{n}\), when \(\nu\) is constant.
Thus, time taken for corf\nuersion decreases with increasing \(n\), with \(\nu\) kept constant.
\(
T \propto \frac{1}{\nu}, \text { when } n \text { remains constant. }
\)
Thus. time taken for con\nuersion decreases with increasing \(\nu\), with \(n\) kept constant.
\(
\Rightarrow \quad T \propto \frac{1}{n \nu}
\)
Thus, time taken for con\nuersion decreases with increase in product of \(n \nu\) and \(T\) is constant. if \(n \nu\) is constant.
A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values \(\lambda_1\), \(\lambda_2\) with \(\lambda_1>\lambda_2\). Which of the following statement are true?
(b, d) According to the question, here given that the de-Broglie wavelength of the particle can be varying cyclically between two values \(\lambda_1\) and \(\lambda_2\), it is possible if the particle is moving in an elliptical orbit with origin as its one focus. As shown in the figure given alongside,
Let \(v_1\) and \(v_2\) be the speed of particle at \(A\) and \(B\) respectively and origin is at focus \(O\). If \(\lambda_1\) and \(\lambda_2\) are the de-Broglie wavelengths associated with particle while moving at \(A\) and \(B\) respectively, then
\(
\lambda_1=\frac{h_1}{m v_1}
\)
and \(\lambda_2=\frac{h}{m v_2}\)
\(
\begin{array}{ll}
\therefore & \frac{\lambda_1}{\lambda_2}=\frac{v_2}{v_1} \\
\text { Since } & \lambda_1>\lambda_2 \\
\therefore & v_2>v_1
\end{array}
\)
By law of conservation of angular momentum, the particle moves faster when it is closer to focus.
From figure, we note that origin \(O\) is closed to \(P\) than \(A\).
Consider a parallel beam of light of wavelength 600 nm and intensity \(100 \mathrm{~W} \mathrm{~m}^{-2}\). (i) Find the energy and linear momentum of each photon. (ii) How many photons cross \(1 \mathrm{~cm}^2\) area perpendicular to the beam in one second?
(a)
(i) The energy of each photon \(E=h c / \lambda\)
\(
=\frac{\left(4.14 \times 10^{-15} \mathrm{eVs}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{600 \times 10^{-9} \mathrm{~m}}=2.07 \mathrm{eV} .
\)
The linear momentum is
\(
p=\frac{E}{c}=\frac{2.07 \mathrm{eV}}{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}=0.69 \times 10^{-8} \mathrm{eVs} \mathrm{~m}^{-1}
\)
(ii) The energy crossing \(1 \mathrm{~cm}^2\) in one second
\(
=\left(100 \mathrm{~W} \mathrm{~m}^{-2}\right) \times\left(1 \mathrm{~cm}^2\right) \times(1 \mathrm{~s})=1.0 \times 10^{-2} \mathrm{~J} .
\)
The number of photons making up this amount of energy is
\(
n=\frac{1.0 \times 10^{-2} \mathrm{~J}}{2.07 \mathrm{eV}}=\frac{1.0 \times 10^{-2}}{2.07 \times 1.6 \times 10^{-19}}=3.0 \times 10^{16}
\)
What is the maximum wavelength of light that can cause photoelectric effect in lithium?
(b) The work function of lithium is 2.5 eV. The threshold wavelength is
\(
\begin{aligned}
\lambda & =h c / \phi \\
& =\frac{\left(4.14 \times 10^{-15} \mathrm{eVs}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{2.5 \mathrm{eV}} \\
& =\frac{1242 \mathrm{eVnm}}{2.5 \mathrm{eV}}=497 \mathrm{~nm}
\end{aligned}
\)
This is the required maximum wavelength.
How many photons are emitted per second by a 5 mW laser source operating at 632.8 nm?
(d) The energy of each photon is
\(
\begin{aligned}
E & =\frac{h c}{\lambda} \\
& =\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(632.8 \times 10^{-9} \mathrm{~m}\right)} \\
& =3.14 \times 10^{-19} \mathrm{~J}
\end{aligned}
\)
The energy of the laser emitted per second is \(5 \times 10^{-3} \mathrm{~J}\). Thus the number of photons emitted per second
\(
=\frac{5 \times 10^{-3} \mathrm{~J}}{3.14 \times 10^{-19} \mathrm{~J}}=1.6 \times 10^{16} .
\)
A monochromatic source of light operating at 200 W emits \(4 \times 10^{20}\) photons per second. Find the wavelength of the light.
(c) The energy of each photon \(=\frac{200 \mathrm{~J} \mathrm{~s}^{-1}}{4 \times 10^{20} \mathrm{~s}^{-1}}[latex] [latex]=5 \times 10^{-19} \mathrm{~J}\).
\(
\begin{aligned}
\text { Wavelength } & =\lambda=\frac{h c}{E} \\
& =\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(5 \times 10^{-19} \mathrm{~J}\right)} \\
& =4.0 \times 10^{-7} \mathrm{~m}=400 \mathrm{~nm}
\end{aligned}
\)
A hydrogen atom moving at a speed \(v\) absorbs a photon of wavelength 122 nm and stops. Find the value of \(v\). Mass of a hydrogen atom \(=1.67 \times 10^{-27} \mathrm{~kg}\).
(b) The linear momentum of the photon
\(
=\frac{h}{\lambda}=\frac{6.63 \times 10^{-34} \mathrm{Js}}{122 \times 10^{-9} \mathrm{~m}}=5.43 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\)
As the photon is absorbed and the atom stops, the total final momentum is zero. From conservation of linear momentum, the initial momentum must be zero. The atom should move opposite to the direction of motion of the photon and they should have the same magnitudes of linear momentum. Thus,
\(
\left(1.67 \times 10^{-27} \mathrm{~kg}\right) v=5.43 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\)
or, \(\quad v=\frac{5.43 \times 10^{-27}}{1.67 \times 10^{-27}} \mathrm{~m} \mathrm{~s}^{-1}=3.25 \mathrm{~m} \mathrm{~s}^{-1}\).
Ultraviolet light of wavelength 280 nm is used in an experiment on photoelectric effect with lithium \((\varphi=2.5 \mathrm{eV})\) cathode. Find (i) the maximum kinetic energy of the photoelectrons and (ii) the stopping potential.
(a)Â
(i) The maximum kinetic energy is
\(
\begin{aligned}
K_{\max } & =\frac{h c}{\lambda}-\varphi \\
& =\frac{1242 \mathrm{eVnm}}{280 \mathrm{~nm}}-2.5 \mathrm{eV} \\
& =4.4 \mathrm{eV}-2.5 \mathrm{eV}=1.9 \mathrm{eV}
\end{aligned}
\)
(ii) Stopping potential \(V\) is given by
\(
e V=K_{\max }
\)
\(
V=\frac{K_{\max }}{e}=\frac{1.9 \mathrm{eV}}{e}=1.9 \mathrm{~V}
\)
A beam of 450 nm light is incident on a metal having work function 2.0 eV and placed in a magnetic field \(B\). The most energetic electrons emitted perpendicular to the field are bent in circular arcs of radius 20 cm. What is the value of \(B\)?
(c) The kinetic energy of the most energetic electrons is
\(
\begin{aligned}
K & =\frac{h c}{\lambda}-\varphi \\
& =\frac{1242 \mathrm{eV} \mathrm{~nm}}{450 \mathrm{~nm}}-2.0 \mathrm{eV} \\
& =0.76 \mathrm{eV}=1.2 \times 10^{-19} \mathrm{~J}
\end{aligned}
\)
The linear momentum \(=m v=\sqrt{2 m \bar{K}}\)
\(
\begin{aligned}
& =\sqrt{2 \times\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \times\left(1.2 \times 10^{-19} \mathrm{~J}\right)} \\
& =4.67 \times 10^{-25} \mathrm{kgms}^{-1}
\end{aligned}
\)
When a charged particle is sent perpendicular to a magnetic field, it goes along a circle of radius
\(
r=\frac{m v}{q B}
\)
Thus, \(0.20 \mathrm{~m}=\frac{4.67 \times 10^{-25} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}}{\left(1.6 \times 10^{-19} \mathrm{C}\right) \times B}\)
\(
B=\frac{4.67 \times 10^{-25} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}}{\left(1.6 \times 10^{-19} \mathrm{C}\right) \times(0.20 \mathrm{~m})}=1.46 \times 10^{-5} \mathrm{~T}
\)
Light described at a place by the equation \(E=\left(100 \mathrm{~V} \mathrm{~m}^{-1}\right)\left[\sin \left(5 \times 10^{15} \mathrm{~s}^{-1}\right) t+\sin \left(8 \times 10^{15} \mathrm{~s}^{-1}\right) t\right]\) falls on a metal surface having work function 2.0 eV . Calculate the maximum kinetic energy of the photoelectrons.
(b) The light contains two different frequencies. The one with larger frequency will cause photoelectrons with largest kinetic energy. This larger frequency is
\(
\begin{aligned}
&v=\frac{\omega}{2 \pi}=\frac{8 \times 10^{15} \mathrm{~s}^{-1}}{2 \pi}\\
&\text { The maximum kinetic energy of the photoelectrons is }\\
&\begin{aligned}
K_{\max } & =h v-\varphi \\
& =\left(4.14 \times 10^{-15} \mathrm{eVs}\right) \times\left(\frac{8 \times 10^{15}}{2 \pi} \mathrm{~s}^{-1}\right)-2.0 \mathrm{eV} \\
& =5.27 \mathrm{eV}-2.0 \mathrm{eV}=3.27 \mathrm{eV}
\end{aligned}
\end{aligned}
\)
Planck constant has the same dimensions as
(d) Force \(\times\) distance \(=\) work done
The unit of work done is Joule.
Therefore the unit of
Force \(\times\) distance \(\times\) time \(=\) Joule \(\times\) second
This is the same unit as Planck’s constant.
Two photons having
(d) The magnitude of momentum and de-Broglie wavelength are related as
\(
\lambda=\frac{h}{p}
\)
Therefore if linear momenta are same then wavelengths are equal but if wavelengths are same then only the magnitudes of momenta are same while direction might be different.
Let \(p\) and \(E\) denote the linear momentum and energy of a photon. If the wavelength is decreased,
(a) If the wavelength of a photon is decreased, both its linear momentum and energy will increase. This is because momentum and energy are inversely proportional to wavelength.
Explanation:
Momentum:
The linear momentum \(p\) of a photon is related to its wavelength \(\lambda\) by the equation \(p=\) \(\mathrm{h} / \lambda\), where h is Planck’s constant. This means that as the wavelength decreases, the momentum increases.
Energy:
The energy \(E\) of a photon is related to its wavelength \(\lambda\) by the equation \(\mathrm{E}=\mathrm{hc} / \lambda\), where \(c\) is the speed of light. Again, this shows that as the wavelength decreases, the energy increases.
Let \(n_r\) and \(n_b\) be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in a given time.
(c)
\(
\begin{aligned}
& P_{\text {Red }}=P_{\text {Blue }} \\
& n_r \frac{h c}{\lambda_r}=n_b \frac{h c}{\lambda_b} \\
& \frac{n_r}{n_b}=\frac{\lambda_r}{\lambda_b} \\
& \text { As } \lambda_r>\lambda_b \\
& \therefore n_r>n_b
\end{aligned}
\)
The equation \(E=p c\) is valid
(c) Energy is given by
\(
\frac{m_0 c^2}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}}
\)
or \(E^2=\frac{m_0^2 c^2}{c^2-v^2}\)
Momentum \(p\) is given by \(p=\frac{m_0 v}{\sqrt{1\left(\frac{v^2}{c^2}\right)}}\)
or \(p^2 c^2=\frac{m_0^2 c^4 v^2}{c^2-v^2}\)
\(
E^2-p^2 c^2=m_0^2 c^4 \text { or } E^2=p^2 c^2+m_0^2 c^4
\)
For photon, rest mass
\(
m_0=0 \text {, so } E=p c, \text { For electron, } m_0 \neq 0 \text {, so } E \neq p c
\)
The work function of a metal is \(h \nu_0\). Light of frequency \(\nu\) falls on this metal. The photoelectric effect will take place only if
(a) Understand the Work Function: The work function \(\phi_0\) of a metal is defined as the minimum energy required to eject an electron from the surface of the metal. It is given by:
\(
\phi_0=h \nu_0
\)
where \(h\) is Planck’s constant and \(\nu_0\) is the threshold frequency.
Energy of Incident Photon: The energy \(\boldsymbol{E}\) of a photon with frequency \(\nu\) is given by:
\(
E=h \nu
\)
Condition for Photoelectric Effect: For the photoelectric effect to occur, the energy of the incident photon must be greater than or equal to the work function of the metal. Thus, we have the condition:
\(
E \geq \Phi_0
\)
Substituting the Expressions: Substituting the expressions for energy and work function into the inequality gives:
\(
h \nu \geq h \nu_0
\)
Cancelling Planck’s Constant: Since \(h\) is a positive constant, we can divide both sides of the inequality by \(h\) without changing the direction of the inequality:
\(
\nu \geq \nu_0
\)
Conclusion: Therefore, the condition for the photoelectric effect to take place is:
\(
\nu \geq \nu_0
\)
Light of wavelength \(\lambda\) falls on a metal having work function \(h c / \lambda_0\). Photoelectric effect will take place only if
(c) The work function, denoted as \(E_0\), is the minimum energy required to eject an electron from the metal surface. This can also be expressed in terms of wavelength as:
\(
E_0=\frac{h c}{\lambda_0}
\)
For the photoelectric effect to occur, the energy of the incoming photon must be greater than or equal to the work function:
\(
E \geq E_0
\)
Substituting the expressions for energy, we have:
\(
\frac{h c}{\lambda} \geq \frac{h c}{\lambda_0}
\)
By simplifying this inequality, we can cancel \(h c\) from both sides (assuming \(h\) and \(c\) are positive constants):
\(
\frac{1}{\lambda} \geq \frac{1}{\lambda_0}
\)
This can be rearranged to:
\(
\lambda \leq \lambda_0
\)
Thus, the condition for the photoelectric effect to take place is that the wavelength of the incident light must be less than or equal to the threshold wavelength of the metal.
When stopping potential is applied in an experiment on photoelectric effect, no photocurrent is observed. This means that
(b) When stopping potential is applied in a photoelectric effect experiment, no photocurrent is observed because the photoelectrons are emitted but are re-absorbed by the emitter metal. In an experiment on photoelectric effect, the photons incident at the metal plate cause photoelectrons to be emitted. The metal plate is termed as “emitter”. The electrons ejected are collected at the other metal plate called “collector”. When the potential of the collector is made negative with respect to the emitter (or the stopping potential is applied), the electrons emitted from the emitter are repelled by the collector. As a result, some electrons go back to the cathode and the current decreases.Â
Explanation:
Stopping potential: This is the minimum voltage applied to the collector plate that completely prevents photoelectrons from reaching it.
Photocurrent: The current created by the flow of photoelectrons.
Process: When light shines on a metal surface, electrons are ejected (photoelectrons). If a stopping potential is applied, the electrons are repelled back to the emitter metal before they reach the collector, thus stopping the photocurrent.
If the frequency of light in a photoelectric experiment is doubled, the stopping potential will
(c) Initial Condition: Let the initial frequency of light be \(\nu\). The stopping potential corresponding to this frequency is \(V_0\). Thus, we can write:
\(
E \cdot V_0=h \nu-\phi_0 \dots(i)
\)
Doubling the Frequency: Now, if the frequency of light is doubled, we have:
\(
\nu^{\prime}=2 \nu
\)
The new stopping potential will be \(V_0^{\prime}\). According to the photoelectric equation, we can write:
\(
E \cdot V_0^{\prime}=h(2 \nu)-\phi_0 \dots(ii)
\)
Substituting and Rearranging: From Equation (ii), we have:
\(
E \cdot V_0^{\prime}=2 h \nu-\phi_0
\)
Now, we can express \(V_0^{\prime}\) in terms of \(V_0\) :
\(
E \cdot V_0^{\prime}=2\left(h \nu-\phi_0\right)+\phi_0
\)
We can substitute \(E \cdot V_0\) from Equation (i) into this equation:
\(
E \cdot V_0^{\prime}=2\left(E \cdot V_0\right)+\phi_0
\)
Dividing the Equations: Now, we can divide Equation (ii) by Equation (i):
\(
\frac{E \cdot V_0^{\prime}}{E \cdot V_0}=\frac{2 h \nu-\phi_0}{h \nu-\phi_0}
\)
Simplifying this gives:
\(
\frac{V_0^{\prime}}{V_0}=\frac{2 h \nu-\phi_0}{h \nu-\phi_0}
\)
Analyzing the Result: We can simplify this further. The right-hand side can be expressed as:
\(
\frac{V_0^{\prime}}{V_0}=2+\frac{\phi_0}{h \nu-\phi_0}
\)
Since \(\phi_0\) is positive and \(h \nu>\phi_0\) (for photoelectric emission to occur), the term \(\frac{\phi_0}{h \nu-\phi_0}\) is also positive.
Conclusion: Therefore, we conclude that:
\(
V_0^{\prime}>2 V_0
\)
This means that the stopping potential when the frequency is doubled is greater than double the original stopping potential.
The frequency and intensity of a light source are both doubled. Consider the following statements.
(A) The saturation photocurrent remains almost the same.
(B) The maximum kinetic energy of the photoelectrons is doubled.
(b) Saturated current varies directly with the intensity of light. As the intensity of light is increased, a large number of photons fall on the metal surface. As a result, a large number of electrons interact with the photons. As a result, the number of emitted electrons increases and, hence, the current also increases.
At the same time, the frequency of the light source also increases.Also, with the increase in frequency of light, the stopping potential increases as well. This will reduce the current. The combined effect of these two is that the current will remain the same
Hence, A is true.
When the frequency of light is doubled from \(\nu\) to \(2 \nu\), it affects the energy of the photons. The energy \(E\) of a photon is given by the equation:
\(
E=h \cdot \nu
\)
Thus, if the frequency is doubled, the energy of each photon becomes:
\(
E^{\prime}=h \cdot(2 \nu)=2 h \cdot \nu
\)
From the Einstein’s photoelectric equation.
Effect on Maximum Kinetic Energy: According to the photoelectric equation, the maximum kinetic energy \(K . E_{\max }\) of the emitted photoelectrons is given by:
\(
K \cdot E_{\max }=h \cdot \nu-\phi
\)
where \(\phi\) is the work function of the material. If the frequency is doubled, the new maximum kinetic energy becomes:
\(
K \cdot E_{\max }^{\prime}=h \cdot(2 \nu)-\phi=2 h \cdot \nu-\phi
\)
This indicates that while the maximum kinetic energy increases, it does not simply double; rather, it increases by an amount dependent on the original frequency and the work function.
Understanding Saturation Photocurrent: The saturation photocurrent is directly related to the number of photoelectrons emitted, which in turn depends on the number of photons striking the surface. The intensity \(I\) of light is defined as:
\(
I=\frac{\Delta(n \cdot E)}{\Delta t \cdot \Delta A}
\)
where \(n\) is the number of photons. If the intensity is doubled while keeping the number of photons constant, the saturation photocurrent remains approximately the same. B is false.
A point source of light is used in a photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential
(c) As the source is removed farther from the emitting metal, the intensity of light will decrease. As the stopping potential does not depend on the intensity of light, it will remain constant.
Explanation: The stopping potential, which is the minimum voltage needed to stop the most energetic electrons from being ejected, depends solely on the frequency of the incident light, not its intensity. Moving the light source farther away reduces the intensity of the light, but it does not change the frequency of the light. Therefore, the stopping potential remains unchanged.
A point source causes photoelectric effect from a small metal plate. Which of the following curves may represent the saturation photocurrent as a function of the distance between the source and the metal?
(d) As the relation between intensity (I) of light and distance (r) is \(I \propto \frac{1}{r^2}\)
As the distance between the source and the metal is increased, it will result in decrease in the intensity of light.
As the saturation current is directly proportional to the intensity of light ( \(i \propto I\) ), it can be concluded that current varies as \(I \propto \frac{1}{r^2}\). Thus, curve d is correct
A nonmonochromatic light is used in an experiment on photoelectric effect. The stopping potential
(c) The photoelectric effect is the ejection of electrons from a metal surface when light radiation falls on it.
The energy of the incoming photon is used for pulling out an electron from the surface and extra energy is converted into the kinetic energy of the electron. So, if a photon has higher energy then the kinetic energy of the electron emitted will be higher. Since non-monochromatic light is used, the electron with maximum kinetic energy will be produced by the component of light with maximum frequency.
This is because the energy of a photon is given as
\(
E=h \nu
\)
where \(h\) is the planck’s constant and \(\nu\) is the frequency of the wave.
The higher the frequency higher will be the energy of the photon.
The photoelectric equation is given as
\(
\Rightarrow h \nu=\phi+\frac{1}{2} m v^2
\)
Where \(\phi\) is the work function and \(\frac{1}{2} m v^2\) is the kinetic energy.
From this, we can write,
\(
\Rightarrow \frac{1}{2} m v^2=h \nu-\phi
\)
In the photoelectric effect, the stopping potential is the potential required for stopping the electrons having maximum kinetic energy. Thus, we can write
\(
\Rightarrow e V_0=h \nu-\phi \dots(i)
\)
Where \(V_o\) is the stopping potential, \(e\) is the charge of an electron.
We know that frequency is related to wavelength by the following relation
\(
c=\nu \lambda
\)
On substituting for \(\nu\) in equation (i) we get
\(
\Rightarrow e V_0=\frac{h c}{\lambda}-\phi
\)
So, we can see that the stopping potential depends on the wavelength.
In the given question non-monochromatic light is used. Which means that light consists of several wavelengths. We know that frequency is inversely related to wavelength. So, the shortest wavelength will have the maximum frequency. Hence it will emit photoelectrons with maximum kinetic energy. Stopping potential would be the potential required to stop these electrons. Therefore, the wavelength \(\lambda\) corresponds to the shortest wavelength. Since kinetic energy is maximum for the shortest wavelength.
A proton and an electron are accelerated by the same potential difference. Let \(\lambda_e\) and \(\lambda_p\) denote the de Broglie wavelengths of the electron and the proton, respectively.
(c) Let \(m_e\) and \(m_p\) be the masses of electron and proton, respectively.
Let the applied potential difference be V.
Thus, the de-Broglie wavelength of the electron,
\(
\lambda_e=\frac{h}{\sqrt{2 m_e e V}} \dots(1)
\)
And de-Broglie wavelength of the proton,
\(
\lambda_p=\frac{h}{\sqrt{2 m_p e V}} \dots(2)
\)
Dividing equation (2) by equation (1), we get :
\(
\begin{aligned}
& \frac{\lambda_p}{\lambda_e}=\frac{\sqrt{m_e}}{\sqrt{m_p}} \\
& m_e<m_p \\
& \therefore \frac{\lambda_p}{\lambda_e}<1 \\
& \Rightarrow \lambda_p<\lambda_e
\end{aligned}
\)
\(
\lambda_e>\lambda_p
\)
When the intensity of a light source is increased,
(a, b)Â When the intensity of a light source in increased, a large number of photons are emitted from the light source. Hence, option (a) is correct.
Due to increase in the number of photons, total energy of the photons emitted per unit time also increases. Hence, option
(b) is also correct.
Increase in the intensity of light increases only the number of photons, not the energy of photons, Hence, option (c) is incorrect.
The speed of photons is not affected by the intensity of light, Hence, option (d) is incorrect.
Photoelectric effect supports quantum nature of light because
(a, b, c)Â Photoelectric effect can be explained on the basis of quantum nature of light. According to the quantum nature of light, energy in light is not uniformly spread. It is contained in packets or quanta known as photons.
Energy of a photon, \({E}={h \nu}\), where h is Planck’s constant and \(\nu\) is the frequency of light.
Above a particular frequency, called threshold frequency, energy of a photon is sufficient to emit an electron from the metal surface and below which, no photoelectron is emitted, as the energy of the photon is low. Hence, option (a) supports the quantum nature of light.
Now, kinetic energy of an electron,
\(
K=h \nu_0-\varphi
\)
Thus, kinetic energy of a photoelectron depends only on the frequency of light (or energy). This shows that if the intensity of light is increased, it only increases the number of photons and not the energy of photons. Kinetic energy of photons can be increased by increasing the frequency of light or by increasing the energy of photon, which supports \({E}={h \nu}\) and, hence, the quantum nature of light. Hence, option (b) also supports the quantum nature of light.
Photoelectrons are emitted from a metal surface even if the metal surface is faintly illuminated; it means that less photons will interact with the electrons. However, few electrons absorb energy from the incident photons and come out from the metal. This shows the quantum nature of light. Hence, (c) also supports the quantum nature of light.
Electric charge of the photoelectrons is quantised; but this statement does not support the quantum nature of light.
A photon of energy \(h f\) is absorbed by a free electron of a metal having work function \(\varphi<h v\).
(d) When light is incident on the metal surface, the photons of light collide with the free electrons. In some cases, a photon can give all the energy to the free electron. If this energy is more then the work-function of the metal,then there are two possibilities. The electron can come out of the metal with kinetic energy \({h \nu}-\varphi\) or it may lose energy on collision with the atoms of the metal and come out with kinetic energy less than \({h \nu}-\varphi\). Thus, it may come out with kinetic energy less than \({h \nu}-\varphi\).
If the wavelength of light in an experiment on photoelectric effect is doubled,
(b, d) For photoelectric effect to be observed, wavelength of incident light should not be more than the largest wavelength called threshold wavelength \(\left(\lambda_0\right)\). If the wavelength of light in an experiment on photoelectric effect is doubled and if it is equal to or less than the threshold wavelength, then photoelectric emission will take place. If it is greater than the threshold wavelength, photoelectric emission will not take place. The photoelectric emission may or may not take place.Photoelectric emission depends on the wavelength of incident light.
Hence, option (b) is correct and (a) is incorrect.
From Einstein’s photoelectric equation,
\(
e V_0=\frac{h c}{\lambda_0}-\varphi,
\)
where \(\mathrm{V}_0=\) stopping potential
\(
\begin{aligned}
& \lambda_0=\text { threshold wavelength } \\
& \mathrm{h}=\text { Planck’s constant } \\
& \varphi=\text { work-function of metal }
\end{aligned}
\)
It is clear that
\(
V_0 \propto \frac{1}{\lambda_0}
\)
Thus, if the wavelength of light in an experiment on photoelectric effect is doubled, its stopping potential will become half.
The photocurrent in an experiment on photoelectric effect increases if
(a) The correct answer is the intensity of the source is increased.
Explanation:
Photocurrent and intensity relationship:
Photocurrent is directly proportional to the intensity of light. A higher intensity means more photons are hitting the metal surface, leading to more electrons being ejected and thus a larger photocurrent.
The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. Light source is put on and a saturation photocurrent is recorded. An electric field is switched on which has a vertically downward direction.
(b) When a downward electric field is applied, the kinetic energy of the electrons will increase because the electric field exerts a force on the electrons, accelerating them in the direction of the field.
As there is no effect of electric field on the number of photons emitted, the photoelectric current will remain same. Hence, option (a) is incorrect.
As the kinetic energy of the electron is increasing, its stopping potential will increase. Hence, option (c) is incorrect.
Threshold wavelength is the characteristic property of the metal and will not change. Hence, (d) is incorrect.
In which of the following situations the heavier of the two particles has smaller de Broglie wavelength? The two particles
(a, c, d) Let \(\mathrm{m}_1\) be the mass of the heavier particle and \(\mathrm{m}_2\) be the mass of the lighter particle.
If both the particles are moving with the same speed v , de Broglie wavelength of the heavier particle,
\(
\lambda_1=\frac{h}{m_1 v} \dots(1)
\)
de Broglie wavelength of the lighter particle,
\(
\lambda_2=\frac{h}{m_2 v} \dots(2)
\)
Thus, from equations (1) and (2), we find that if the particles are moving with the same speed v , then \(\lambda_1<\lambda_2\).
Hence, option (a) is correct.
If they are moving with the same linear momentum, then using the de Broglie relation \(\lambda=\frac{h}{p}\)
We find that both the bodies will have the same wavelength. Hence, option (b) is incorrect.
If K is the kinetic energy of both the particles, then de Broglie wavelength of the heavier particle,
\(
\lambda_1=\frac{h}{\sqrt{2 m_1 K}}
\)
de Broglie wavelength of the lighter particle,
\(
\lambda_2=\frac{h}{\sqrt{2 m_2 K}}
\)
It is clear from the above equation that if \(m_1>m_2\), then \(\lambda_1<\lambda_2\).
Hence, option (c) is correct.
When they have fallen through the same height h , then velocity of both the bodies,
\(
v=\sqrt{2 g h}
\)
Now,
\(
\begin{aligned}
& \lambda_1=\frac{h}{m_1 \sqrt{2 g h}} \\
& \lambda_2=\frac{h}{m_2 \sqrt{2 g h}} \\
& m_1>m_2 \\
& \therefore \lambda_1<\lambda_2
\end{aligned}
\)
Hence, option (d) is correct.
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
(a) Maximum kinetic energy of photoelectrons \(K_{\max }=e V_0\) where, \(e=1.6 \times 10^{-19} \mathrm{C}, V_0=1.5 \mathrm{~V}\)
\(
\therefore \quad K_{\max }=1.6 \times 10^{-19} \times 1.5=2.4 \times 10^{-19} \mathrm{~J}
\)
In an experimental study of photoelectric effect, the stopping potential for a metallic surface is 5.68 V. What will be the maximum velocity of the electrons?
(b) Stopping potential, \(V_0=5.68 \mathrm{~V}, m_e=9.1 \times 10^{-31} \mathrm{~kg}\)
\(
\begin{aligned}
& \text { Maximum KE, } K_{\max }=\frac{1}{2} m v_{\max }^2=e V_0 \\
& \Rightarrow v_{\max }=\sqrt{\frac{2 e V_0}{m}}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 5.68}{9.1 \times 10^{-31}}}=1.414 \times 10^6 \mathrm{~ms}^{-1}
\end{aligned}
\)
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm ?
(b) Here, \(W=\) Work function of certain metal \(=4.2 \mathrm{eV}\)
\(
\begin{aligned}
=\left(4.2 \times 1.6 \times 10^{-19}\right) \mathrm{J}=6.72 \times 10^{-19} \mathrm{~J} & \\
& \left(\because 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)
\end{aligned}
\)
If \(E\) be the energy of the photon of incident light, then
\(
E=h \nu=\frac{h c}{\lambda}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{330 \times 10^{-9}} \mathrm{~J}=6.01 \times 10^{-19} \mathrm{~J}
\)
The photoelectric emission is possible only if \(E=h \nu>\phi\), where \(\phi\) is work function of metal. Here \(E<\phi\), so this metal will not give photoelectric emission for incident radiation of wavelength 330 nm.
The threshold frequency for a certain metal is \(3.3 \times 10^{14} \mathrm{~Hz}\). If light of frequency \(8.2 \times 10^{14} \mathrm{~Hz}\) is incident on the metal, predict the cut-off voltage for the photoelectric emission.
(d) Using the relation, \(h \nu-h \nu_0=e V_0\)
We get \(V_0=h\left(\frac{\nu-\nu_0}{e}\right)\)
Here, \(\nu_0=3.3 \times 10^{14} \mathrm{~Hz}, \nu=8.2 \times 10^{14} \mathrm{~Hz}\)
\(
=6.62 \times 10^{-34} \times\left(\frac{8.2 \times 10^{14}-3.3 \times 10^{14}}{1.6 \times 10^{-19}}\right)=2.03 \mathrm{~V}
\)
Light of frequency \(7.21 \times 10^{14} \mathrm{~Hz}\) is incident on a metal surface. Electrons with a maximum speed of \(6.0 \times 10^5 \mathrm{~m} / \mathrm{s}\) are ejected from the surface. What is the threshold frequency for photoemission of electrons?
(a) Using Einstein’s photoelectric equation, \(h \nu-h \nu_0=\frac{1}{2} m v_{\max }^2\),
We get \(h \nu_0=h \nu-\frac{1}{2} m v_{\text {max }}^2\)
\(
\therefore \quad \nu_0=\nu-\frac{m v_{\max }^2}{2 h}
\)
Here, \(\quad \nu=7.21 \times 10^{14} \mathrm{~Hz}\)
\(
v_{\max }=6.0 \times 10^5 \mathrm{~m} / \mathrm{s}
\)
and \(\quad m=\) mass of electron \(=9.1 \times 10^{-31} \mathrm{~kg}\)
\(\therefore\) Threshold frequency, \(\nu_0\)
\(
\begin{aligned}
& =7.21 \times 10^{14}-\frac{9.1 \times 10^{-31} \times\left(6.0 \times 10^5\right)^2}{2 \times 6.62 \times 10^{-34}} \\
& =7.21 \times 10^{14}-2.47 \times 10^{14}=4.74 \times 10^{14} \mathrm{~Hz}
\end{aligned}
\)
The photoelectric work function of potassium is 2.3 eV. If light having a wavelength of \(2800 Ã…\) falls on potassium, find
(i) the kinetic energy in electron volts of the most energetic electrons ejected
(ii) and the stopping potential in volts.
(a) Given, \(W=2.3 \mathrm{eV}\) and \(\lambda=2800 Ã…\)
\(
\therefore E(\text { in } \mathrm{eV})=\frac{h c}{\lambda(\text { in } Ã…)}=\frac{12375}{\lambda(\text { in } Ã…)}=\frac{12375}{2800}=4.4 \mathrm{eV}
\)
(i) As \(K_{\text {max }}=E-W=(4.4-2.3) \mathrm{eV}=2.1 \mathrm{eV}\)
(ii) Using the relation, \(K_{\text {max }}=e V_0\)
\(
\therefore \quad 2.1 \mathrm{eV}=e V_0 \text { or } V_0=2.1 \mathrm{~V}
\)
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the cathode is made.
(c) Energy of incident photon,
\(
\begin{aligned}
E & =\frac{h c}{\lambda}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}} \\
& =4.08 \times 10^{-19} \mathrm{~J}
\end{aligned}
\)
Using formula \(e V_0=E-W\), we get
\(
\begin{aligned}
W & =E-e V_0 \\
E & =h \nu
\end{aligned}
\)
and \(\quad W=\) Work function of the material
\(
\begin{aligned}
& \text { Here, } E=4.08 \times 10^{-19} \mathrm{~J} \\
& e=1.6 \times 10^{-19} \mathrm{C} \\
& \text { and } V_0=0.38 \mathrm{~V} \\
& \therefore \quad W
\end{aligned}
\)
The work function of cesium is 2.14 eV. Calculate
(i) the threshold frequency for cesium and
(ii) the wavelength of the incident light, if the photocurrent is brought to zero by a stopping potential of 0.60 V.
Given, \(h=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}\).
(b) Here, \(V_0=0.60 \mathrm{~V}\)
\(
\phi_0=2.14 \mathrm{eV}=2.14 \times 1.6 \times 10^{-19} \mathrm{~J}
\)
(i) Threshold frequency, \(\nu_0=\frac{\phi_0}{h} \quad\left(\because \phi_0=h \nu_0\right)\)
\(
=\frac{2.14 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}=5.16 \times 10^{14} \mathrm{~Hz}
\)
(ii)
\(
\text { As, } \begin{aligned}
e V_0 & =\frac{h c}{\lambda}-\phi_0 \text { or } \lambda=\frac{h c}{\left(e V_0+\phi_0\right)} \\
& =\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{\left(1.6 \times 10^{-19} \times 0.6+2.14 \times 1.6 \times 10^{-19}\right)} \\
& \approx 454 \times 10^{-9} \mathrm{~m}=454 \mathrm{~nm}
\end{aligned}
\)
The work function of cesium metal is 2.14 eV. When light of frequency \(6 \times 10^{14} \mathrm{~Hz}\) is incident on the metal surface, photoemission of electrons occurs. What is the
(i) maximum kinetic energy of the emitted electrons,
(ii) stopping potential
(iii) and maximum speed of the emitted photoelectrons?
(i) Using Einstein’s photoelectric equation,
\(
\begin{aligned}
E_{\max } & =h \nu-\phi_0 \\
& =6.62 \times 6 \times 10^{-20}-2.14 \times 1.6 \times 10^{-19} \\
& =39.72 \times 10^{-20}-34.24 \times 10^{-20} \mathrm{~J} \\
& =5.48 \times 10^{-20} \mathrm{~J} \\
& =\frac{0.548 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}\left(\because 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right) \\
& =0.342 \mathrm{eV}
\end{aligned}
\)
(ii) As, \(e V_0=E_{\max }\)
\(
V_0=\frac{E_{\max }}{e}=\frac{0.342 \times 1.6 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{C}}
\)
\(
\begin{aligned}
&=0.342 \mathrm{JC}^{-1}=0.342 \mathrm{~V}\\
&\text { (iii) As we know that, } \frac{1}{2} m v_{\text {max }}^2=E_{\text {max }}\\
&\therefore v_{\max }=\sqrt{\frac{2 E_{\max }}{m}}=\sqrt{\frac{2 \times 5.48 \times 10^{-20}}{9.1 \times 10^{-31}}}=3.45 \times 10^5 \mathrm{~ms}^{-1}
\end{aligned}
\)
When light of wavelength \(4000 Ã…\) is incident on a barium emitter, emitted photoelectrons in a transverse magnetic field strength \(B=5.2 \times 10^{-6} T\) moves in a circular path. Find the radius of the circle. (Take, work function of barium is 2.5 eV and mass of electron is \(9.1 \times 10^{-31} \mathrm{~kg}, e=1.6 \times 10^{-19} \mathrm{C}\) and \(h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\) )
(d)
\(
\begin{aligned}
&\text { Using the relation, } K_{\max }=E-\phi_0\\
&\begin{aligned}
& \quad \frac{1}{2} m v_{\max }^2=\frac{h c}{\lambda}-\phi_0 \Rightarrow v_{\max }^2=\frac{2}{m}\left[\frac{h c}{\lambda}-\phi_0\right] \\
& =\frac{2}{9.1 \times 10^{-31}}\left[\frac{\left(6.6 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{4 \times 10^{-7}}-2.5 \times 1.6 \times 10^{-19}\right] \\
& =20.88 \times 10^{10} \\
& v_{\max }=4.57 \times 10^5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\end{aligned}
\)
Radius of circular path described by emitted photoelectrons in a magnetic field is given by
\(
\begin{aligned}
\quad B e v_{\max } & =\frac{m v_{\max }^2}{r} \\
\Rightarrow \quad r & =\frac{m v_{\max }}{B e}=\frac{\left(9.1 \times 10^{-31}\right) \times\left(4.57 \times 10^5\right)}{5.2 \times 10^{-6} \times 1.6 \times 10^{-19}}=\frac{1}{2} \mathrm{~m}
\end{aligned}
\)
Light of wavelength \(2475 Ã…\) is incident on barium. Photoelectrons emitted describe a circle of radius 100 cm by a magnetic field of flux density \(\frac{1}{\sqrt{17}} \times 10^{-5} \mathrm{~T}\). Find the work function of the barium. \(\left(\right.\) Take, \(\left.\frac{e}{m}=1.7 \times 10^{11}\right)\)
(c) Radius of circular path described by a charged particle in a magnetic field \(B\) is given by
\(
r=\frac{\sqrt{2 m K}}{q B} \Rightarrow K=\frac{q^2 B^2 r^2}{2 m}=\left(\frac{e}{m}\right) \frac{e B^2 r^2}{2}
\)
where, \(K=\) kinetic energy of a particle,
\(
q=\text { charge on a particle }
\)
and \(r=\) radius of circular path.
So,
\(
\begin{aligned}
K & =\frac{1}{2} \times 1.7 \times 10^{11} \times 1.6 \times 10^{-19} \times\left(\frac{1}{\sqrt{17}} \times 10^{-5}\right) \times(1)^2 \\
& =8 \times 10^{-20} \mathrm{~J}=0.5 \mathrm{eV}
\end{aligned}
\)
Using the relation, \(E=\phi_0+K_{\text {max }}\)
\(
\Rightarrow \quad \phi_0=E-K_{\max }
\)
where,
\(
\begin{aligned}
E & =h \nu=\frac{h c}{\lambda}=\frac{12375}{\lambda(\operatorname{in} \AA)} \\
& \phi_0=\left(\frac{12375}{2475}\right) \mathrm{eV}-0.5 \mathrm{eV}=4.5 \mathrm{eV}
\end{aligned}
\)
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be \(4.12 \times 10^{-15} \mathrm{Vs}\). Calculate the value of Planck’s constant.
(b) We know, \(e V_0=h \nu-W\)
\(
\begin{array}{ll}
\therefore & V_0=\left(\frac{h}{e}\right) \nu-\frac{W}{e} \\
\text { Given, } & \text { Slope }=4.12 \times 10^{-15} \mathrm{Vs}=\frac{h}{e} \\
\therefore & h
\end{array}
\)
A student performs an experiment on photoelectric effect, using two materials \(A\) and \(B\). A plot of \(V_{\text {stop }}\) versus v is given in the figure.
(i) Which material A or B has a higher work function?
(ii) Given the electric charge of an electron \(=1.6 \times 10^{-19} \mathrm{C}\), find the value of \(h\) obtained from the experiment for both \(A\) and \(B\). Comment on whether it is consistent with the Einstein’s theory.
(b)
(i) Given, threshold frequency of \(A\) is given by \(\nu_{O A}=5 \times 10^{14} \mathrm{~Hz}\) and for \(B, \quad \nu_{O B}=10 \times 10^{14} \mathrm{~Hz}\)
We know that, work function, \(\phi=h \nu_0\) or \(\phi_0 \propto \nu_0\)
So, \(\quad \frac{\phi_{O A}}{\phi_{O B}}=\frac{5 \times 10^{14}}{10 \times 10^{14}}<1 \quad \Rightarrow \phi_{O A}<\phi_{O B}\)
Thus, work function of \(B\) is higher than \(A\).
(ii) For metal \(A\), slope \(=\frac{h}{e}=\frac{2}{(10-5) \times 10^{14}}\)
\(
\begin{aligned}
h & =\frac{2 \times e}{5 \times 10^{14}}=\frac{2 \times 1.6 \times 10^{-19}}{5 \times 10^{14}} \\
& =6.4 \times 10^{-34} \mathrm{Js}
\end{aligned}
\)
For metal \(B\), slope \(=\frac{h}{e}=\frac{2.5}{(15-10) \times 10^{14}}\)
\(
\begin{aligned}
h & =\frac{2.5 \times e}{5 \times 10^{14}}=\frac{2.5 \times 1.6 \times 10^{-19}}{5 \times 10^{14}} \\
& =8 \times 10^{-34} \mathrm{Js}
\end{aligned}
\)
Since the value of \(h\) from experiment for metals \(A\) and \(B\) is different. Hence, experiment is not consistent with the Einstein’s theory.
Which of the following statement(s) is correct?
(c) Both (a) and (b).
Explanation:
Hertz discovered the photoelectric effect:
German physicist Heinrich Hertz first observed the phenomenon of photoelectric emission in 1887 while working on experiments with radio waves. He noticed that when ultraviolet light shone on metal electrodes, it caused a change in voltage.
Lenard studied the photoelectric effect in detail:
Philipp Lenard, another German physicist, conducted further research on the photoelectric effect after Hertz’s initial discovery. He investigated the properties of the emitted electrons, including their kinetic energy and dependence on the frequency and intensity of the incident light.
When intensity of light increases, then photoelectric current
(a)Â When the intensity of light increases, the photoelectric current increases.
Explanation:
Photoelectric effect:
This phenomenon occurs when light strikes a metal surface and electrons are ejected from it.
Intensity and number of photons:
Higher intensity light means more photons are hitting the metal surface per unit time.
More photons, more electrons:
Each photon can eject one electron, so with more photons incident, there are more electrons ejected, resulting in a higher photoelectric current.
The number of photoelectrons emitted per second from a metal surface increases when
(d) The intensity of the incident light increases.
Explanation:
Concept: The number of photoelectrons emitted per second is directly proportional to the intensity of the incident light.
Reasoning: Higher intensity means more photons hitting the metal surface, leading to a greater number of electrons being ejected.
If the work function of potassium is 2 eV, then its photoelectric threshold wavelength is
\(
\begin{aligned}
&\text { (b) } \phi_0=h \nu=\frac{h c}{\lambda_0} \text {, where } \phi_0=2 \mathrm{eV} \text {. }\\
&\therefore \lambda(\text { in } Ã…)=\frac{12375}{2}=6187.5 \approx 6200 Ã…=620 \mathrm{~nm}
\end{aligned}
\)
Maximum kinetic energy of emitted electron from the surface of a metal is \(K\). When frequency of incident light is doubled, then maximum kinetic energy of emitted electrons is \(K^{\prime}\) then
\(
\begin{aligned}
(a) \text { As, } & K_{\max } =E-W \\
\Rightarrow & K =E-W \dots(i) \\
\text { and } & K^{\prime} =2 E-W
\end{aligned}
\)
\(
\begin{aligned}
& =2(E)-(E-K) \quad \text { [using Eq. (i)] } \\
& =E+K=K+W+K=W+2 K \\
K^{\prime} & >2 K (\because W>0)
\end{aligned}
\)
When frequency increases, slope of stopping potential versus frequency graph
(c) When frequency increases, the slope of a stopping potential versus frequency graph remains the same.
Explanation: According to the photoelectric equation, the stopping potential is directly proportional to the frequency of incident light, meaning the slope of the graph representing this relationship is a constant value \((h/e)\), where \(h\) is Planck’s constant and \(e\) is the charge of an electron.
Key points:
Photoelectric equation:
VO \(=(h / e)^{\nu}-\phi_0\), where \(V_O\) is stopping potential, \(h\) is Planck’s constant, \(\nu\) is frequency, and \(\)\phi_0\(\) is the work function.
Constant slope: Since the equation is linear with a constant coefficient (h/e), the slope remains the same regardless of the frequency increase.
The work function of aluminium is 4.2 eV. If two photons, each of energy 3.5 eV strike on an electron of aluminium, then emission of electrons will be
(b) For photoemission to take place, \(h \nu>\phi_0\). Here, \(E<\phi_0\), so it is not possible.
The momentum of photon of electromagnetic radiation is \(3.3 \times 10^{-29} \mathrm{~kg}-\mathrm{ms}^{-1}\). Find out the frequency and wavelength of the wave associated with it.
(c)Â Given, \(h=6.63 \times 10^{-34} \mathrm{Js}, c=3 \times 10^8 \mathrm{~ms}^{-1}\)
\(
p=3.3 \times 10^{-29} \mathrm{~kg}-\mathrm{ms}^{-1}
\)
(i) Momentum,
\(
\begin{aligned}
p & =\frac{h \nu}{c} \Rightarrow \nu=\frac{p c}{h}=\frac{3.3 \times 10^{-29} \times 3 \times 10^8}{6.63 \times 10^{-34}} \\
& =1.5 \times 10^{13} \mathrm{~Hz}
\end{aligned}
\)
(ii) \(\quad \lambda=\frac{c}{\nu}=\frac{3 \times 10^8}{1.5 \times 10^{13}}=2 \times 10^{-5} \mathrm{~m}\)
The wavelength of a photon is \(1.4 Ã…\). It collides with an electron. The energy of the scattered electron is \(4.26 \times 10^{-16} \mathrm{~J}\). Find the wavelength of photon after collision. (Take, \(h=6.63 \times 10^{-34} \mathrm{Js}\) )
(d) Let initial wavelength of photon is \(\lambda_1\) and final wavelength of photon is \(\lambda_2\).
Since, total energy and momentum are conserved in photon-electron collision.
\(
\begin{aligned}
& \Rightarrow 4.26 \times 10^{-16}=\frac{h c}{\lambda_1}-\frac{h c}{\lambda_2} \\
& \quad \frac{h c}{\lambda_2}=\frac{h c}{\lambda_1}-4.26 \times 10^{-16} \\
& \Rightarrow \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{\lambda_2}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.4 \times 10^{-10}}-4.26 \times 10^{-16}
\end{aligned}
\)
\(
\begin{aligned}
& =14.20 \times 10^{-16}-4.26 \times 10^{-16} \\
\lambda_2 & \approx 2 \times 10^{-10} \mathrm{~m} \\
& =2.0 Ã…
\end{aligned}
\)
The eye can detect \(5 \times 10^4\) photons per square metre per second of green light \((\lambda=5000 Ã…)\) while the ear can detect \(10^{-13}\left(\mathrm{~W} / \mathrm{m}^2\right)\). Find the factor by which the eye is more sensitive as a power detector than the ear.
(c)
\(
\begin{aligned}
& \text { As, } E=\frac{h c}{\lambda(Ã…)} \\
& \Rightarrow \quad E=\frac{12375}{5000}=2.475 \mathrm{eV}=4 \times 10^{-19} \mathrm{~J} \\
& \begin{aligned}
I_{\text {Eye }} & =(\text { Photon flux }) \times(\text { Energy of photon }) \\
& =\left(5 \times 10^4\right) \times\left(4 \times 10^{-19}\right)=2 \times 10^{-14} \mathrm{~W} / \mathrm{m}^2 \\
& \frac{S_{\text {Eye }}}{S_{\text {Ear }}}=\frac{I_{\text {Ear }}}{I_{\text {Eye }}}=\frac{10^{-13}}{2 \times 10^{-14}}=5
\end{aligned}
\end{aligned}
\)
As, lesser the intensity, more sensitive is the detector therefore Eye is 5 times more sensitive than the ear.
The intensity of sunlight reaching the surface of the earth is \(1.388 \times 10^3 \mathrm{~W} / \mathrm{m}^2\). How many photons (nearly) per square metre are incident on the earth per second? Assume that, the photons in the sunlight have an average wavelength of 550 nm.
(b) Let, \(n=\) number of photons incident on earth per sec per \(\mathrm{m}^2\) and \(E^{\prime}=\) energy of each photon \(=\frac{h c}{\lambda}\).
If \(E\) be the energy of all photons reaching the surface of earth per sec per unit area, then
\(
E=n E^{\prime}=\frac{n h c}{\lambda}
\)
Intensity = Energy of photons reaching the earth per second per unit area
\(
\begin{aligned}
\text { Given, } \quad I & =1.388 \times 10^3 \mathrm{~W} / \mathrm{m}^2 \\
& =1.388 \times 10^3 \mathrm{Js}^{-1} \mathrm{~m}^{-2} \\
\therefore \quad \frac{n h c}{\lambda} & =I \\
\text { or } \quad n & =\frac{I}{\left(\frac{h c}{\lambda}\right)}=\frac{1.388 \times 10^3}{h c} \times \lambda \\
& =\frac{1.388 \times 10^3 \times 550 \times 10^{-9}}{6.62 \times 10^{-34} \times 3 \times 10^8} \\
& =3.84 \times 10^{21}
\end{aligned}
\)
A \(100 W\) sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (i) What is the energy per photon associated with the sodium light? (ii) At what rate are the photons delivered to the sphere?
(a)
(i) Using the formula, \(E=\frac{h c}{\lambda}\), we get
\(
\begin{aligned}
E & =\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{589 \times 10^{-9}} \\
& =3.37 \times 10^{-19} \mathrm{~J} \\
& =\frac{3.37 \times 10^{-19}}{1.6 \times 10^{-19}}=2.1 \mathrm{eV}
\end{aligned}
\)
(ii) Let \(n=\) rate at which the photons are delivered to the sphere.
\(
\begin{aligned}
n & =\frac{\text { Energy radiated per second }}{\text { Energy of each photon }}=\frac{P}{E} \\
& =\frac{100}{3.38 \times 10^{-19}}=2.96 \times 10^{20} \text { photons } / \mathrm{s}
\end{aligned}
\)
(i) Find the momentum of a photon of light of wavelength 660 nm.
(ii) Calculate the number of photons emitted per second by a 66 W source of monochromatic light of wavelength 660 nm .
(iii) How many photons per second does a one watt bulb emit, if its efficiency is 20\% and the wavelength of light emitted is 400 nm?
(iv) Monochromatic light of wavelength 200 nm is incident normally on a surface of area \(2 \mathrm{~cm}^2\). If the intensity of light is \(100 \mathrm{~W} / \mathrm{m}^2\), calculate the rate at which photons strike the surface.
(a)
(i) Momentum of a photon,
\(
\begin{aligned}
p & =\frac{h}{\lambda}=\frac{6.6 \times 10^{-34}}{660 \times 10^{-9}} \\
& =1.00 \times 10^{-27} \mathrm{~kg}-\mathrm{m} / \mathrm{s}
\end{aligned}
\)
(ii) Power of the source, \(P=\frac{n h c}{\lambda}\)
\(
\therefore \quad n=\frac{P \lambda}{h c}=\frac{66 \times 660 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8}=2.2 \times 10^{20}
\)
(iii) Efficiency \(=\frac{\text { Output power }}{\text { Input power }}\)
\(
\begin{array}{ll}
\Rightarrow & \frac{20}{100}=\frac{P}{1} \\
\Rightarrow & P=\frac{1}{5} \mathrm{~W}
\end{array}
\)
\(\therefore\) Number of photons emitted,
\(
\begin{aligned}
n & =\frac{P \lambda}{h c}=\frac{1}{5} \times \frac{400 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} \\
& =0.04 \times 10^{19}=4 \times 10^{17}
\end{aligned}
\)
\(
\begin{aligned}
&\text { (iv) Rate at which photons strike the surface }\\
&\begin{aligned}
n & =\frac{P \lambda}{h c}=\frac{I A \lambda}{h c}=\frac{100 \times 2 \times 10^{-4} \times 200 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} \\
& =2 \times 10^{16} \text { per second }
\end{aligned}
\end{aligned}
\)
The rest mass of photon is
(a) Actually rest mass of the photon is zero. But it’s effective mass is given as
\(
\begin{aligned}
& E=m c^2=h \nu \\
& \Rightarrow m=\frac{E}{c^2}=\frac{h \nu}{c^2}=\frac{h}{c \lambda}
\end{aligned}
\)
This mass is also known as the kinetic mass of the photon
Photons are electrically neutral and therefore, are not deflected by magnetic or electrical fields.
The rest mass of a photon is \(\mathbf{0}\). Because light always travels with some speed and is never at rest. So it has zero rest mass.
Irrespective of the intensity, all the photons which have the same frequency, possess the same energy.
With the increase in intensity, the number of photons emitted also increases. Also, the total momentum and energy remain conserved during a collision between a photon and particle.
The momentum of the photon of wavelength \(5000 Ã…\) will be
\(
\begin{aligned}
&\text { (a) Momentum, }\\
&\begin{aligned}
p & =\frac{h}{\lambda}=\frac{6.6 \times 10^{-34}}{\left(5000 \times 10^{-10}\right)} \\
& =1.3 \times 10^{-27} \mathrm{~kg}-\mathrm{ms}^{-1}
\end{aligned}
\end{aligned}
\)
The momentum of a photon is \(2 \times 10^{-16} \mathrm{~g}-\mathrm{cms}^{-1}\). Its energy is
\(
\begin{aligned}
&\text { (c) Momentum, } p=\frac{E}{c}\\
&\Rightarrow \quad E=p c=2 \times 10^{-16} \times\left(3 \times 10^8\right)=6 \times 10^{-8} \mathrm{erg}
\end{aligned}
\)
Amongst the following relation, which of one is correct?
 (a) Momentum,Â
\(
\begin{aligned}
p & =\frac{E}{c} \\
E & =p c \Rightarrow E^2=p^2 c^2
\end{aligned}
\)
The number of photons of wavelength 540 nm emitted per second by an electric bulb of power 100 W is (Take, \(h=6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\) )
\(
\begin{aligned}
&\text { (c) Power, } P=\frac{n h c}{\lambda}\\
&\therefore \quad 100=\frac{n \times 6 \times 10^{-34} \times 3 \times 10^8}{540 \times 10^{-9}} \Rightarrow n=3 \times 10^{20}
\end{aligned}
\)
Light of intensity \(10^4 \mathrm{~W} / \mathrm{m}^2\) is falling on a perfectly reflecting surface. Pressure on the surface is
(a) Pressure \(=\frac{2 I}{c}\) (For perfectly reflecting surface) where, \(c\) is speed of light.
\(
\begin{aligned}
& =\frac{2 \times 10^4}{3 \times 10^8}=\frac{2}{3} \times 10^{-4} \mathrm{~N} / \mathrm{m}^2 \\
& =0.667 \times 10^{-4}=6.67 \times 10^{-5} \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)
Light of intensity \(I_0\), exerts a pressure of \(4.5 \times 10^{-5} \mathrm{~N} / \mathrm{m}^2\) on a perfectly absorbing surface, Value of \(I_0\) is
\(
\begin{aligned}
& \text { (b) Pressure }=\frac{I_0}{c} \text { (For a perfectly absorbing surface) } \\
& \Rightarrow 4.5 \times 10^{-5}=\frac{I_0}{3 \times 10^8} \\
& \Rightarrow \quad I_0=13.5 \times 10^3=1.35 \times 10^4 \simeq 1.4 \times 10^4 \mathrm{~W} / \mathrm{m}^2
\end{aligned}
\)
A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of \(3 \times 10^6 \mathrm{~ms}^{-1}\). What is the velocity of the particle?
(b) de-Broglie wavelength,
\(
\lambda=\frac{h}{m v}
\)
As both particle and electron are having same wavelength, therefore their momentum will be equal.
\(
\begin{aligned}
& & m_p v_p & =m_e v_e \\
\Rightarrow & & v_p & =\frac{m_e v_e}{m_p}=\frac{9.1 \times 10^{-31} \times 3 \times 10^6}{10^{-6}} \\
\Rightarrow & & v_p & =2.7 \times 10^{-18} \mathrm{~ms}^{-1}
\end{aligned}
\)
When the momentum of a proton is changed by an amount \(p_0\), the corresponding change in the de-Broglie wavelength is found to be \(0.25 \%\). Find the original momentum of the photon.
(a) As \(\lambda \propto \frac{1}{p} \Rightarrow \frac{\Delta p}{p}=-\frac{\Delta \lambda}{\lambda}\)
\(
\Rightarrow \quad \frac{p_0}{p}=\frac{0.25}{100} \Rightarrow p=400 p_0
\)
The kinetic energy of an electron is 5 eV. Calculate the de-Broglie wavelength associated with it (Take, \(h=6.6 \times 10^{-34} \mathrm{Js}\) and \(m_e=9.1 \times 10^{-31} \mathrm{~kg}\) )
(d) de-Broglie wavelength associated with the electron,
\(
\begin{aligned}
\lambda & =\frac{h}{\sqrt{2 m(\mathrm{KE})}} \\
& =\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 5 \times 1.6 \times 10^{-19}}} \\
& =5.469 \times 10^{-10} \mathrm{~m}=5.47 Ã…
\end{aligned}
\)
Find out the energy that should be added to an electron in terms of initial energy to reduce its de-Broglie wavelengths from \(10^{-10} \mathrm{~m}\) to \(0.5 \times 10^{-10} \mathrm{~m}\).
(a) de-Broglie wavelength is given by
\(
\begin{aligned}
\lambda & =\frac{h}{\sqrt{2 m E}} \Rightarrow \lambda \propto \frac{1}{\sqrt{E}} \\
\frac{\lambda_1}{\lambda_2} & =\sqrt{\frac{E_2}{E_1}} \Rightarrow \frac{10^{-10}}{0.5 \times 10^{-10}}=\sqrt{\frac{E_1}{E_2}} \\
\Rightarrow \quad E_2 & =4 E_1
\end{aligned}
\)
Hence, added energy \(=E_2-E_1=3 E_1\)
What is the
(i) momentum,
(ii) speed and
(iii) de-Broglie wavelength of an electron with kinetic energy of 120 eV?
(b) (i) Using the relation, \(E=\frac{p^2}{2 m}\),
\(
\text { We get } \quad \begin{aligned}
p & =\sqrt{2 m E} \\
& =\sqrt{2 \times 9.1 \times 10^{-31} \times 120 \times 1.6 \times 10^{-19}} \\
& =5.91 \times 10^{-24} \mathrm{~kg}-\mathrm{ms}^{-1}
\end{aligned}
\)
(ii) Speed, \(v=\frac{p}{m}=\frac{5.91 \times 10^{-24}}{9.1 \times 10^{-31}}=6.5 \times 10^6 \mathrm{~ms}^{-1}\)
(iii) de-Broglie wavelength is given by
\(
\lambda=\frac{h}{p}=\frac{6.62 \times 10^{-34}}{5.91 \times 10^{-24}}=1.12 \times 10^{-10} \mathrm{~m}
\)
What is the de-Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that, the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen \(=14.0076 u\) )
(d) Let \(v=\mathrm{rms}\) speed of \(\mathrm{N}_2\) molecule at 300 K
\(
v=\sqrt{\frac{3 k T}{m}}
\)
where, \(k=\) Boltzmann constant \(=1.38 \times 10^{-23} \mathrm{JK}^{-1}\)
\(
\begin{aligned}
v & =\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{28.0152 \times 1.67 \times 10^{-27}}} \\
& =5.15 \times 10^2 \mathrm{~ms}^{-1}
\end{aligned}
\)
Using the formula, \(\lambda=\frac{h}{m v}\), we get
\(
\begin{aligned}
\lambda & =\frac{6.62 \times 10^{-34}}{28.0152 \times 1.67 \times 10^{-27} \times 5.15 \times 10^2} \\
& =2.75 \times 10^{-11} \mathrm{~m}
\end{aligned}
\)
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(i) an electron and
(ii) a neutron, would have the same de-Broglie wavelength?
(c) (i) Let \(E_1=\mathrm{KE}\) of electron
Using the relation, \(\lambda=\frac{h}{\sqrt{2 m_e E}}\), we get
\(
\begin{aligned}
\lambda^2 & =\frac{h^2}{2 m_e E} \\
\therefore \quad E_1 & =\frac{h^2}{2 m_e \lambda^2}=\frac{\left(6.62 \times 10^{-34}\right)^2}{2 \times 9.1 \times 10^{-31} \times\left(589 \times 10^{-9}\right)^2} \\
& =6.94 \times 10^{-25} \mathrm{~J} \\
& =\frac{6.94 \times 10^{-25}}{1.6 \times 10^{-19}} \mathrm{eV}=4.34 \times 10^{-6} \mathrm{eV}
\end{aligned}
\)
\(
\begin{aligned}
&\text { (ii) } E_2=\mathrm{KE} \text { of neutron }\\
&\begin{aligned}
\therefore E_2 & =\frac{h^2}{2 m_n \lambda^2}=\frac{\left(6.62 \times 10^{-34}\right)^2}{2 \times 1.67 \times 10^{-27} \times\left(589 \times 10^{-9}\right)^2} \\
& =3.782 \times 10^{-28} \mathrm{~J} \\
& =\frac{3.782 \times 10^{-28}}{1.6 \times 10^{-19}} \mathrm{eV}=2.36 \times 10^{-9} \mathrm{eV}
\end{aligned}
\end{aligned}
\)
(i) For which value of kinetic energy of a neutron will the associated de-Broglie wavelength be \(1.40 \times 10^{-10} \mathrm{~m}\) ?
(ii) Also find the de-Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of \(\frac{3}{2} k T\) at 300 K.
(a) (i) Using the formula \(\lambda=\frac{h}{\sqrt{2 m E}}\), we get
\(
\begin{aligned}
& \lambda^2=\frac{h^2}{2 m E} \\
E= & \frac{h^2}{2 m \lambda^2}=\frac{\left(6.62 \times 10^{-34}\right)^2}{2 \times 1.67 \times 10^{-27} \times\left(1.40 \times 10^{-10}\right)^2} \\
= & 6.69 \times 10^{-21} \mathrm{~J} \\
= & \frac{6.69 \times 10^{-21}}{1.6 \times 10^{-19}} \mathrm{eV} \\
= & 4.184 \times 10^{-2} \mathrm{eV}
\end{aligned}
\)
(ii) Here, \(T=\) absolute temperature \(=300 \mathrm{~K}\)
\(
\begin{aligned}
k & =\text { Boltzmann constant } \\
& =1.38 \times 10^{-23} \mathrm{JK}^{-1} \\
E & =\text { kinetic energy of neutron } \\
& =\frac{3}{2} k T=\frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \\
& =6.21 \times 10^{-21} \mathrm{~J}
\end{aligned}
\)
Using the relation, \(\lambda=\frac{h}{\sqrt{2 m E}}\), we get
\(
\begin{aligned}
\lambda & =\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 6.21 \times 10^{-21}}} \\
& =1.455 \times 10^{-10} \mathrm{~m}=1.455 Ã…
\end{aligned}
\)
Who confirmed that, the intensity of scattered beam of electrons was not the same but different at different angles of scattering?
(b) The wave nature of the material particles as predicted by de-Broglie, was confirmed by CI Davisson and LH Germer (1927) in United States and independently by GP Thomson (1928) in Scotland.
In Davisson and Germer experiment, electron beam of wavelength \(1.5 Ã…\) is incident on a crystal of lattice spacing \(3 Ã…\). Find the incident angle for first maxima.
(c) It is given that, \(d=3 Ã…, \lambda=1.5 Ã…\) and \(n=1\)
For the crystal, we can write, \(2 d \sin \theta=n \lambda\)
\(
\begin{array}{lc}
\Rightarrow & 2 \times 3 \times \sin \theta=1 \times 1.5 \\
\Rightarrow & \sin \theta=\frac{1}{4} \\
\Rightarrow & \theta=14.4^{\circ}
\end{array}
\)
\(\therefore\) Angle of incidence of electron beams, \(\theta=14.4^{\circ}\)
In Davisson and Germer experiment, second maxima is obtained for an electron beam of wavelength \(2.5 Ã…\) and incident at an angle of \(30^{\circ}\). Evaluate the lattice spacing of the crystal.
(a) It is given that, wavelength of electron beam, \(\lambda=2.5 Ã…\)
Angle of incidence, \(\theta=30^{\circ}\)
\(n=2\) (for second maxima), \(d=\) ?
As we know that, \(2 d \sin \theta=n \lambda\)
\(
\begin{array}{ll}
\Rightarrow & 2 d \sin 30^{\circ}=2 \times 2.5 Ã… \\
\Rightarrow & 2 \times d \times \frac{1}{2}=2 \times \frac{5}{2} \\
\Rightarrow & d=5 Ã…
\end{array}
\)
\(\therefore\) Lattice spacing, \(d=5 Ã…\).
An electron microscope is used to study very minute objects, such as viruses and microbes. etc. can have magnification in the order of
(b) An electron microscope can have a very high magnification of \(\approx 10^5\).
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
(c) Here, \(V=50 \mathrm{kV}=5 \times 10^4 \mathrm{~V}, m_e=9.11 \times 10^{-31} \mathrm{~kg}\)
KE of an electron,
\(
K=50 \mathrm{eV}=\left(1.6 \times 10^{-19} \times 5 \times 10^4\right) \mathrm{J}=8 \times 10^{-15} \mathrm{~J}
\)
\(\therefore\) de-Broglie wavelength of electrons,
\(
\begin{aligned}
\lambda & =\frac{h}{\sqrt{2 m K}}=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 8 \times 10^{-15}}} \mathrm{~m} \\
& =\frac{6.63 \times 10^{-11}}{12.07}=5.5 \times 10^{-12} \mathrm{~m}
\end{aligned}
\)
Wavelength of yellow light, \(\lambda_y=5.9 \times 10^{-7} \mathrm{~m}\)
Resolving power of a microscope \(\propto 1 / \lambda\)
\(
\begin{aligned}
& \therefore \frac{\text { Resolving power of electron microscope }}{\text { Resolving power of optical microscope }}=\frac{\lambda_y}{\lambda} \\
& \qquad=\frac{5.9 \times 10^{-7}}{5.5 \times 10^{-12}} \equiv 10^5
\end{aligned}
\)
Thus, the resolving power of an electron microscope is about \(10^5\) times greater than that of an optical microscope.
The wavelength of the matter wave is independent of which amongst the given associated variable/physical quantity?
(d) Charge. The wavelength of a matter wave is independent of the charge of the particle.
Explanation:
The wavelength of a matter wave is determined by its momentum, which is given by the equation \(p=m v\) where \(m\) is mass and \(v\) is velocity. Since the equation for wavelength is \(\lambda=\frac{h}{p}\), the wavelength is directly dependent on mass and velocity, but not on charge.
Why other options are incorrect:
(a) Mass: The wavelength is dependent on mass because it is part of the momentum equation.
(b) Velocity: The wavelength is dependent on velocity as it is part of the momentum equation.
(c) Momentum: The wavelength is directly dependent on momentum, making it not independent of this quantity.
If the velocity of an electron is doubled, its de-Broglie frequency will
(c) As, \(\lambda=\frac{h}{m v}\)
where, \(v=\) velocity of an electron
and \(\lambda=\) de-Broglie wavelength.
Also, \(\lambda \propto 1 / \nu\) (de-Broglie frequency)
When \(v\) is doubled, \(\lambda\) gets halved and \(\nu\) (de-Broglie frequency) gets doubled.
Proton and \(\alpha\)-particles have the same de-Broglie wavelength. Which quantity will remain same for both of them?
\(
\begin{aligned}
&\text { (d) As, } \lambda=\frac{h}{p} \text {, so for } \lambda_p=\lambda_\alpha \text {, we get }\\
&p_p=p_\alpha
\end{aligned}
\)
A photon, an electron and a very small unknown particle have the same wavelength. The one with the most energy is
(a) All the three particles have same wavelength, therefore they will same momentum say \(p\).
Energy of photon \(E=p c\). Therefore, energy of photon is maximum.
Explanation: The photon has the most energy. When a photon, an electron, and a small unknown particle have the same wavelength, the photon will have the most energy because its energy is directly proportional to its momentum, which is inversely proportional to its wavelength. Since all three particles have the same wavelength, and the photon has no mass and travels at the speed of light, it will have the highest energy.
If \(\lambda_p\) and \(\lambda_e\) denote the de-Broglie wavelength of proton and electron after they are accelerated from rest through the same potential difference, then
\(
\begin{array}{lll}
\text { (c) } \lambda=\frac{h}{\sqrt{2 q V m}} & \text { or } & \lambda \propto \frac{1}{\sqrt{m}} \\
\text { As, } &Â m_e<m_p \\
\therefore & \lambda_e>\lambda_p
\end{array}
\)
When the kinetic energy of an electron is increased, the wavelength of the associated wave will
\(
\begin{array}{ll}
\text { (b) Wavelength, } \lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}} & \\
\therefore \quad \lambda \propto \frac{1}{\sqrt{E}} & \text { ( } h \text { and } m \text { are constants) }
\end{array}
\)
So, when the KE of an electron is increased, the wavelength of the associated wave will decrease.
The de-Broglie wavelength associated with an electron, accelerated through a potential difference of 100 V is
(a) The de-Broglie wavelength, \(\lambda=\frac{1.227}{\sqrt{V}} \mathrm{~nm}\) Here, \(V=100 \mathrm{~V}\)
\(
\Rightarrow \quad \lambda=\frac{1.227}{\sqrt{100}} \mathrm{~nm} \simeq 0.123 \mathrm{~nm}
\)
The de-Broglie wavelength of a neutron at \(27^{\circ} \mathrm{C}\) is \(\lambda\). What will be its wavelength at \(927^{\circ} \mathrm{C}\)?
(a) As, \(\lambda \propto \frac{1}{\sqrt{T}} \Rightarrow \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{T_2}{T_1}}\)
Here, \(T_2=927^{\circ} \mathrm{C}=(927+273) \mathrm{K}=1200 \mathrm{~K}\)
\(
\begin{aligned}
T_1 & =27^{\circ} \mathrm{C}=(27+273) \mathrm{K}=300 \mathrm{~K} \\
\lambda_1 & =\lambda \\
\therefore \quad \frac{\lambda}{\lambda_2} & =\sqrt{\frac{1200}{300}} \Rightarrow \lambda_2=\frac{\lambda}{2}
\end{aligned}
\)
Which amongst the following property is exhibited by matter waves?
(d) Diffraction of the matter waves.
Diffraction: Matter waves, like light waves, can bend around corners and pass through narrow openings, demonstrating wave-like behavior, a key property that can be observed in experiments like the double-slit experiment.
In Davisson and Germer experiment, intensity of scattered beam of electron was found to be maximum when angle of scattering is
(a) In the Davisson and Germer experiment, the intensity of the scattered beam of electrons was found to be maximum when the angle of scattering was \(50^{\circ}\).
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