Past NEET Entrance Papers

Hint

(a) Let \(m_e\) and \(m_{\alpha}\) be the masses of electron and alpha particle, 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 alpha particle,
\(
\lambda_{\alpha}=\frac{h}{\sqrt{2 m_{\alpha} e V}} \dots(2)
\)
Dividing equation (2) by equation (1), we get :
\(
\begin{aligned}
& \frac{\lambda_{\alpha}}{\lambda_e}=\frac{\sqrt{m_e}}{\sqrt{m_{\alpha}}} \\
& m_e<m_{\alpha} \\
& \therefore \frac{\lambda_{\alpha}}{\lambda_e}<1 \\
& \Rightarrow \lambda_{\alpha}<\lambda_e
\end{aligned}
\)

Hint

(c) Work function of the material: \(\phi \mathrm{eV}\)
Wavelength of incident light: \(\lambda=\frac{h c}{{e}} \mathrm{~m}\)
Plank’s constant: \({h}\)
Speed of light: \(c\)
Electron charge: \(e\)
Energy of a photon: \(E=\frac{h c}{\lambda}\)
Photoelectric effect equation: \({K}_{\max }={E}-{\phi}\), where \(\boldsymbol{K}_{\max }\) is the maximum kinetic energy of the emitted electron, \({E}\) is the energy of the incident photon, and \(\phi\) is the work function of the material.

How to solve?
Calculate the energy of the incident photon and then use the photoelectric effect equation to find the maximum kinetic energy of the emitted electron.

Step 1: Calculate the energy of the incident photon
The energy of the incident photon is given by:
\(E=\frac{h c}{\lambda}\)

Substitute the given value of \(\lambda\) :
\(E=\frac{h c}{\frac{h c}{e}}\)
\(E=e\)

Step 2: Calculate the maximum kinetic energy of the photo-electron
Use the photoelectric effect equation:
\(K_{\max }=E-\phi\)

Substitute the calculated value of \(\boldsymbol{E}\) :
\(K_{\max }=e-\phi\)

The maximum kinetic energy of the photo-electron is \(e-\phi\).

Hint

(d) de Broglie wavelength formula: \(\lambda=\frac{\boldsymbol{h}}{\boldsymbol{p}}\) Kinetic energy formula: \(E=\frac{p^2}{2 m}\)

How to solve?
Express the kinetic energy in terms of the de Broglie wavelength and then analyze the relationship.

Step 1: Express momentum in terms of de Broglie wavelength
From the de Broglie wavelength formula:
\(p=\frac{h}{\lambda}\)

Step 2: Substitute momentum into the kinetic energy formula
Substitute \(p\) in the kinetic energy formula:
\(E=\frac{\left(\frac{h}{\lambda}\right)^2}{2 m}\)
\(E=\frac{h^2}{2 m \lambda^2}\)

Step 3: Express \(\frac{1}{\lambda^2}\) in terms of \(E\)
Rearrange the equation to isolate \(\frac{1}{\lambda^2}\) :
\(\frac{1}{\lambda^2}=\frac{2 m E}{h^2}\)

Step 4: Analyze the relationship
\(\frac{1}{\lambda^2}\) is directly proportional to \(E\) :
\(\frac{1}{\lambda^2} \propto E\)

The graph showing the variation of \(\frac{1}{\lambda^2}\) and \(E\) is a straight line passing through the origin.

Hint

(b) Photons are massless particles.
Photons are electrically neutral.
Energy and momentum are always conserved in a closed system.

Step 1: Evaluate statement A
The energy of a photon is given by \({E}={h} {\nu}\), where \({h}\) is Planck’s constant and \(\nu\) is the frequency.
This statement is correct.

Step 2: Evaluate statement B
Photons are massless particles that always travel at the speed of light \(c\) in vacuum. This statement is correct.

Step 3: Evaluate statement C
The momentum of a photon is given by \(p=\frac{{E}}{{c}}\).
Since \(E=h \nu\), then \(p=\frac{h \nu}{c}\).
This statement is correct.

Step 4: Evaluate statement D
In any collision, including photon-electron collisions, both total energy and total momentum are conserved, according to the laws of physics.
This statement is correct.

Step 5: Evaluate statement E
Photons are electrically neutral particles.
This statement is incorrect.

Step 6: Determine the correct answer

Statements A, B, C, and D are correct.
Statement E is incorrect.

Hint

(c) Wavelength of radiation: \(\lambda=280 \mathrm{~nm}=280 \times 10^{-9} \mathrm{~m}\)
Work function of the cathode: \(\phi=2.5 \mathrm{eV}\)
Planck’s constant: \(h=6.62 \times 10^{-34} \mathrm{~J} \mathrm{~s}\)
Speed of light: \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
\(
1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}
\)
Energy of a photon: \(E=h f=\frac{h c}{\lambda}\)
Photoelectric effect equation: \(K_{\max }={E}-{\phi}\)
How to solve
Calculate the energy of the incident photon and then subtract the work function to find the maximum kinetic energy of the photoelectrons.

Step 1: Calculate the energy of the photon
The energy of a photon is given by \(E=\frac{h c}{\lambda}\).
Substituting the given values:
\(E=\frac{\left(6.62 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{280 \times 10^{-9} \mathrm{~m}}\)
\(E=7.09 \times 10^{-19} \mathrm{~J}\)

Step 2: Convert the photon energy to electron volts
To convert Joules to electron volts, divide by \(1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\) :
\(E=\frac{7.09 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\)
\(E=4.43 \mathrm{eV}\)

Step 3: Calculate the maximum kinetic energy
The maximum kinetic energy of the photoelectrons is given by \(K_{\max }={E}-{\phi}\).
Substituting the values:
\(K_{\text {max }}=4.43 \mathrm{eV}-2.5 \mathrm{eV}\)
\(K_{\max }=1.93 \mathrm{eV}\)

The maximum kinetic energy of the photoelectrons is 1.93 eV.

Hint

(d) Statement I is incorrect, as the de Broglie wavelength depends on the particle’s momentum (mass and velocity), not its charge or nature. Statement II is correct, as the wave nature of particles in the sub-atomic domain is a fundamental concept and has been experimentally verified.

Here’s a more detailed explanation:
De Broglie Wavelength:
The de Broglie wavelength \((\lambda)\) of a particle is related to its momentum (p) by the equation \(\lambda=h / p\), where \(h\) is Planck’s constant.
Momentum (p): Momentum is a measure of an object’s mass in motion and is calculated as \({p}={mv}\) (mass x velocity).
Charge and Nature: The de Broglie wavelength is not dependent on the charge or nature of the particle. Any moving particle, regardless of its charge or type, possesses a wavelength.

Wave Nature in Sub-atomic Domain:
The wave-particle duality of matter is a cornerstone of quantum mechanics. Subatomic particles, like electrons, exhibit both wave-like and particle-like behavior.
Experimental Evidence: The Davisson-Germer experiment provided strong evidence for the wave nature of electrons by demonstrating their diffraction, a phenomenon characteristic of waves.
Measurability: The wave nature of subatomic particles is not just a theoretical concept; it’s a measurable phenomenon with practical applications in areas like electron microscopy and quantum computing.

Hint

(d) To find the de Broglie wavelength associated with an electron accelerated by a potential difference, we use the formula:
\(
\lambda=\frac{h}{\sqrt{2 m e V}}
\)
where \(h\) is Planck’s constant, \(m\) is the mass of the electron, \(e\) is the charge of the electron, and \(V\) is the potential difference.
Here given, \(V=81 \mathrm{~V} . h=6.626 \times 10^{-34} \mathrm{Js} . m=9.11 \times 10^{-31} \mathrm{~kg} . e=1.6 \times 10^{-19} \mathrm{C}\).
\(
\lambda=\frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 81}}=0.136 \mathrm{~nm}
\)

Short-cut method:

\(
\lambda=\frac{12.27}{\sqrt{V}}=\frac{12.27}{\sqrt{81}}==0.136 {~nm}
\)

Hint

(b) According to Einstein’s photoelectric equation:
\(
\begin{aligned}
& {E}=\varphi+{KE}_{\max } \\
& {E}={h} \nu \\
& {E}=\varphi+{KE}_{\max }={h} \nu
\end{aligned}
\)
\(
K E_{\max }=(h \nu-\varphi)
\)
Thus maximum kinetic energy depends on the frequency (\(\)\nu\(\)) of incident radiation.
The maximum kinetic energy doesn’t depend upon the intensity of incident radiations and number of photons.
Since speed of radiation is always constant and equal to speed of light. So maximum kinetic energy is independent of the speed of the incident radiations. 

Hint

(a)

\(
\text { We know that } e^{-} \text {emitts when } h \nu>\phi_0
\)
Here it is clear that energy of photon is more than the work function of Cs [Caesium] only so Ans. only (Cs).

Explanation:
Work function of Caesium (Cs): \(W_{C s}=2.14 \mathrm{eV}\)
Work function of Potassium \((\mathrm{K}): W_K=2.30 \mathrm{eV}\)
Work function of Sodium (Na): \(W_{N a}=2.75 \mathrm{eV}\)
Incident photon energy: \(E=2.20 \mathrm{eV}\)

Photoelectric emission occurs when the incident photon energy is greater than the work function of the metal.

How to solve?
Compare the incident photon energy with the work functions of each metal.

Step 1: Compare the incident energy with the work function of Caesium
Check if \(E>W_{C S}\) :
\(2.20 \mathrm{eV}>2.14 \mathrm{eV}\)
Photoemission occurs for Caesium.

Step 2: Compare the incident energy with the work function of Potassium
Check if \({E}>{W}_{{K}}\) :
\(2.20 \mathrm{eV}<2.30 \mathrm{eV}\)
Photoemission does not occur for Potassium.

Step 3: Compare the incident energy with the work function of Sodium
Check if \(E>W_{N a}\) :
\(2.20 \mathrm{eV}<2.75 \mathrm{eV}\)
Photoemission does not occur for Sodium.

Only Caesium will emit photoelectrons.

Hint

(a) By the law of photo-electric effect \(\frac{h c}{\lambda}=e V\) or \(\lambda=\frac{h c}{e V} \propto \frac{1}{V}\)

Explanation:

An electron is accelerated through a potential difference of \(V\) volts.

The energy of an electron accelerated through a potential difference \(V\) is \(e V\).
The energy of a photon is \(E=h f=\frac{h c}{{\lambda}}[latex].
The minimum wavelength corresponds to the maximum energy of the photon.
How to solve?
Equate the electron’s kinetic energy to the photon’s energy and solve for the wavelength.

Step 1: Equate the electron’s kinetic energy to the photon’s energy
The kinetic energy gained by the electron is eV .
The energy of the emitted photon is [latex]\frac{h c}{\lambda_{\text {min }}}\).
Equating the two energies:
\(\mathrm{eV}=\frac{h c}{\lambda_{\text {min }}}\)

Step 2: Solve for \(\lambda_{\text {min }}\)
Rearrange the equation to solve for \(\lambda_{\text {min }}\) :
\(\lambda_{\text {min }}=\frac{h c}{e V}\)

Step 3: Determine the proportionality
From the equation \(\lambda_{\min }=\frac{h c}{e V^{\prime}}\), where \(h, c\), and \(e\) are constants, we can see that:
\(\lambda_{\min } \propto \frac{1}{V}\)

The minimum wavelength of X-rays is inversely proportional to the potential difference \(V\).

Hint

(c)

\(
\begin{aligned}
&\text { We know, } K E_{\max }=h f-h f_0\\
&e V_0=h f-h f_0\left(\because K E_{\max }=e V_0\right)\\
&e V_0=4.2 \mathrm{eV}-2.2 \mathrm{eV}\\
&\therefore V_0=2 V
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
&\text { According to Einstein’s photoelectric equation }\\
&\begin{aligned}
& (K . E)_{\max }=h f-h f_0 \\
& (K . E)_{\max }=h\left(4 f_0\right)-h f_0 \\
& (K . E)_{\max }=3 h f_0
\end{aligned}
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
&\text { Since } k_{\max }=e V_s=h f-\phi\\
&\begin{aligned}
& \frac{e V_s}{2}=h f-h f_0 \ldots \ldots \text { (i) } \\
& e V_s=\frac{h f}{2}-h f_0 \ldots \ldots \text { (ii) } \\
& \frac{1}{2}\left[\frac{h f}{2}-h f_0\right]=h f-h f_0 \\
& \Rightarrow h f_0-\frac{h f_0}{2}=h f-\frac{h f}{4} \\
& \Rightarrow \frac{h f_0}{2}=\frac{3 h f}{4} \\
& f_0=\frac{3 f}{2}
\end{aligned}
\end{aligned}
\)

Hint

(a) According to the de-Broglie hypothesis according to which all matters have both particle and wave nature.

According to de Broglie’s hypothesis, \(\lambda={h} / {p} \dots(1)\)
where \({h}=\) plank’s constant, \({p}=\) momentum, \(\lambda=\) wavelength of particle.
Equation (1) shows that \(\lambda \propto 1 / p\), which can be related to \(y=1 / x\) graph
This type of graph is called a rectangular hyperbola.
As \(\lambda \propto 1 / p\) we will have a rectangular hyperbola.

Hint

(c) Initial temperature: \(T_1=27^{\circ} \mathrm{C}\)
Final temperature: \({T}_{\mathbf{2}}=927^{\circ} \mathrm{C}\)
Initial de-Broglie wavelength: \(\lambda_1=\lambda\)

Convert Celsius to Kelvin using: \(\left.{T}(\mathrm{K})={T}\left({ }^{\circ} \mathrm{C}\right)+273.15 \approx {T}{ }^{\circ} \mathrm{C}\right)+273\)
de-Broglie wavelength is inversely proportional to the square root of temperature:
\(
\lambda \propto \frac{1}{\sqrt{T}}
\)
How to solve?
Use the relationship between de-Broglie wavelength and temperature to find the new wavelength.

Step 1: Convert temperatures from Celsius to Kelvin.
\(
\begin{aligned}
& T_1=27+273=300 \mathrm{~K} \\
& T_2=927+273=1200 \mathrm{~K}
\end{aligned}
\)

Step 2: Set up the ratio of de-Broglie wavelengths.
Since \(\lambda \propto \frac{1}{\sqrt{T}}\), we have \(\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{T_2}{T_1}}\)

Step 3: Substitute the given values and solve for \(\lambda_2\).
\(
\begin{aligned}
& \frac{\lambda}{\lambda_2}=\sqrt{\frac{1200}{300}} \\
& \frac{\lambda}{\lambda_2}=\sqrt{4} \\
& \frac{\lambda}{\lambda_2}=2 \\
& \lambda_2=\frac{\lambda}{2}
\end{aligned}
\)
The de-Broglie wavelength at \(927^{\circ} \mathrm{C}\) is \(\frac{\lambda}{2}\).

Hint

(c) Since blue light can eject photoelectrons but green light cannot, and the frequency of light determines its ability to eject them, then violet light, which has a higher frequency than blue light, would also be capable of ejecting photoelectrons.

Here’s a more detailed explanation:
Photoelectric Effect: The photoelectric effect occurs when light shines on a metal surface, and electrons are emitted. The energy of the light (determined by its frequency) must be greater than or equal to a certain threshold energy (specific to the metal) for electrons to be ejected.

Frequency and Energy: The frequency of light is directly related to its energy. Higher-frequency light has higher energy.

Blue Light Ejects, Green Light Doesn’t: The fact that blue light ejects electrons but green light doesn’t mean that the frequency of blue light is above the metal’s threshold frequency, while the frequency of green light is below it.

Violet Light’s Higher Frequency: Violet light has a higher frequency than blue light, meaning it has even more energy.
Conclusion:
Because violet light has a higher frequency (and therefore higher energy) than blue light, it will also be able to eject photoelectrons from the same metal.

Hint

(a) The energy of a single photon is given by:
\(E=h \frac{c}{\lambda}\)
\(E=\left(6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s}\right) \times \frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{600 \times 10^{-9} \mathrm{~m}}\)
\(E=\frac{19.8 \times 10^{-26}}{600 \times 10^{-9}} \mathrm{~J}\)
\(E=3.3 \times 10^{-19} \mathrm{~J}\)

Calculate the number of photons emitted per second.
The number of photons emitted per second \((n)\) is the total power divided by the energy of a single photon:
\({n}=\frac{{P}}{{E}}\)
\(n=\frac{3.3 \times 10^{-3} \mathrm{~W}}{3.3 \times 10^{-19} \mathrm{~J}}\)
\(n=10^{16}\) photons/s

The number of photons emitted per second is \(10^{16}\).

Hint

(a)

\(
\begin{aligned}
& K_{\max }=\frac{h c}{\lambda} \\
& \Rightarrow \frac{p^2}{2 m}=\frac{h c}{\lambda} \\
& \Rightarrow p=\sqrt{\frac{2 m h c}{\lambda}} \\
& \Rightarrow \lambda_d=\frac{h}{p}=h \sqrt{\frac{\lambda}{2 m h c}}=\sqrt{\frac{\lambda h}{2 m c}} \\
& \Rightarrow \lambda=\left(\frac{2 m c}{h}\right) \lambda_d^2
\end{aligned}
\)

Hint

(c) Given, Initial frequency of light \(f=1.5 f_0\), where \(f_0\) is the threshold frequency.
New frequency \(f^{\prime}=\frac{f}{2}\).
New intensity \(I^{\prime}=2 I\), where \(I\) is the initial intensity.

Photoelectric effect occurs only if the frequency of incident light is greater than the threshold frequency.
The photoelectric current is proportional to the intensity of the incident light when the frequency is above the threshold frequency.
The formula for the maximum kinetic energy of emitted electrons is \(K E_{\max }=h f-\phi\), where \(h\) is Planck’s constant and \(\phi\) is the work function.
The work function \(\phi\) is related to the threshold frequency by \(\phi=h f_0\).

How to solve?
Determine if the new frequency is above or below the threshold frequency and then determine the effect on the photoelectric current.

Step 1: Calculate the new frequency.
The new frequency is half of the initial frequency:
\(f^{\prime}=\frac{1.5 f_0}{2}=0.75 f_0\)
Step 2: Compare the new frequency with the threshold frequency.
The new frequency \(f^{\prime}=0.75 f_0\) is less than the threshold frequency \(f_0\).
Step 3: Determine if photoelectric emission occurs.
Since the new frequency is less than the threshold frequency, no photoelectrons will be emitted.
Step 4: Determine the photoelectric current.
Because no photoelectrons are emitted, the photoelectric current will be zero, regardless of the intensity.
The photoelectric current becomes zero.

Hint

(c) How to solve?
Equate the de Broglie wavelength formula with the kinetic energy formula and solve for \(V\).

Step 1: Express the de Broglie wavelength in terms of kinetic energy.
\(
\lambda=\frac{h}{\sqrt{2 m_e K E}}
\)

Step 2: Express the kinetic energy in terms of potential difference.
\(
K E=e V
\)

Step 3: Substitute \(K E\) in the de Broglie wavelength formula.
\(
\lambda=\frac{h}{\sqrt{2 m_e e V}}
\)

Step 4: Solve for \(V\).
\(
V=\frac{h^2}{2 m_e e \lambda^2}
\)

Step 5: Substitute the given values and calculate \({V}\).
\(
\begin{aligned}
& V=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)^2}{2 \times\left(9.109 \times 10^{-31} \mathrm{~kg}\right) \times\left(1.602 \times 10^{-19} \mathrm{C}\right) \times\left(1.227 \times 10^{-11} \mathrm{~m}\right)^2} \\
& V \approx 10^4 \mathrm{~V}
\end{aligned}
\)
The potential difference is approximately \(10^4 \mathrm{~V}\).

Hint

(a) 

\(
\begin{aligned}
& \lambda=\frac{12.27}{\sqrt{V}} Å \\
& =\frac{12.27}{\sqrt{144} \times 10^{10}} \\
& =1.02 \times 10^{-10} \mathrm{~m} \\
& =102 \times 10^{-3} \mathrm{~nm}
\end{aligned}
\)

Hint

(c) For an electron accelerated through a potential \(V\), \(\lambda=\frac{12.27}{\sqrt{V}} \AA=\frac{12.27 \times 10^{-10}}{\sqrt{10000}}=12.27 \times 10^{-12} \mathrm{~m}\)

Hint

(d)

\(
\begin{aligned}
& \phi=\frac{h c}{\lambda} \\
& \Rightarrow 4 \mathrm{eV}=\frac{1240}{\lambda} \\
& \Rightarrow=\frac{1240}{4}=310 \mathrm{~nm}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& \lambda=\frac{h}{\sqrt{2 m E_k}}\left(\because E_k \text { is same for both }\right) \\
& \Rightarrow \frac{\lambda_p}{\lambda_a}=\sqrt{\frac{4 m}{m}} \\
& =2: 1
\end{aligned}
\)

Hint

(a) Initial de-Broglie wavelength
\(
\lambda_0=\frac{h}{m V_0} \dots(i)
\)
Acceleration of electron
\(
\mathrm{a}=\frac{e E_0}{m} \quad\left(\because \mathrm{~F}=m a=e E_0\right)
\)
Velocity after time ‘ \(t\) ‘
\(
\mathrm{V}=\left(V_0+\frac{e E_0}{m} t\right)
\)
So, \(\lambda=\frac{h}{m V}=\frac{h}{m\left(V_0+\frac{e E_0}{m} t\right)}\)
\(
=\frac{h}{m V_0\left[1+\frac{e E_0}{m V_0} t\right]}=\frac{\lambda_0}{\left[1+\frac{e E_0}{m V_0} t\right]} \dots(ii)
\)
Dividing eqs. (ii) by (i),
de-Broglie wavelength \(\lambda=\frac{\lambda_0}{\left[1+\frac{e E_0}{m V_0} t\right]}\)

Hint

(a) Given, Threshold frequency \(=\nu_0\)
Frequency of incident light \(=2 {\nu}_0\)
\(
\begin{aligned}
& h\left(2 \nu_0\right)=h \nu_0+\frac{1}{2} m v_1^2 \\
& h \nu_0=\frac{1}{2} m v_1^2 \dots(i) \\
& 4 h \nu_0=\frac{1}{2} m v_2^2 \dots(ii)
\end{aligned}
\)
Divide (i) by (ii),
\(
\begin{aligned}
& \frac{1}{4}=\frac{v_1^2}{v_2^2} \\
& \frac{v_1}{v_2}=\frac{1}{2}
\end{aligned}
\)

Hint

(a) Kinetic energy of a neutron in thermal equilibrium with heavy water at a temperature \(T\) is given as
\(
K=\frac{3}{2} k T \dots(i)
\)
Also momentum ( \(p\) ) is, \(p=\sqrt{2 m K}\)
From eqn. (i)
\(
p=\sqrt{2 m \cdot \frac{3}{2} k T}=\sqrt{3 m k T}
\)
Required de-Broglie wavelength is given as
\(
\lambda=\frac{h}{p}=\frac{h}{\sqrt{3 m k T}}
\)

Hint

(a) The maximum kinetic energy is given as
\(
K_{\max }=h \nu-\phi_0=h \nu-h \nu_o=\frac{h c}{\lambda}-\frac{h c}{\lambda_0}
\)
where \(\lambda_0=\) threshold wavelength
or \(\frac{1}{2} m v^2=\frac{h c}{\lambda}-\frac{h c}{\lambda_0}\)
Here, \(h=4.14 \times 10^{-15} \mathrm{eV} \mathrm{s}, c=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\)
\(
\begin{aligned}
& \lambda_o= 3250 \times 10^{-10} \mathrm{~m}=3250 Å \\
& \lambda=2536 \times 10^{-10} \mathrm{~m}=2536 Å, \\
& m=9.1 \times 10^{-31} \mathrm{~kg} \\
& h c=4.14 \times 10^{-15} \mathrm{eV} \mathrm{~s} \times 3 \times 10^8 \mathrm{~ms}^{-1} \\
&=12420 \mathrm{eV} \mathrm{~A} \\
& \therefore \quad \frac{1}{2} m v^2=12420\left[\frac{1}{2536}-\frac{1}{3250}\right] \mathrm{eV} \\
&= 1.076 \mathrm{eV} \\
& v^2=\frac{2.152 \mathrm{eV}}{m}=\frac{2.152 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \\
& \therefore \quad v \approx 6 \times 10^5 \mathrm{~ms}^{-1}=0.6 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
\)

Hint

(a) Kinetic energy of electrons
\(
K=\frac{p^2}{2 m}=\frac{(h / \lambda)^2}{2 m}=\frac{h^2}{2 m \lambda^2}
\)
So, maximum energy of photon \(=K\)
\(
\frac{h c}{\lambda_0}=\frac{h^2}{2 m \lambda^2} \quad \therefore \lambda_0=\frac{2 m c \lambda^2}{h}
\)

Hint

(d) According to Einstein’s photoelectric equation maximum kinetic energy of photoelectrons,
\(
K E_{\max }=E-\phi
\)
or \(2=5-\phi \quad \therefore \phi=3 \mathrm{eV}\)
When \(E=6 \mathrm{eV}\) then
\(
\begin{aligned}
& K E_{\max }=6-3=3 \mathrm{eV} \\
& e\left(V_{\text {cathode }}-V_{\text {anode }}\right)=3 \mathrm{eV}
\end{aligned}
\)
The stopping potential is defined relative to the cathode. Since the anode is at a higher potential than the cathode, the stopping potential relative to the anode ( \(V_s\)) is:
\(
V_{s}=-3 \mathrm{~V}
\)

Hint

(d) For electron of energy \(E\), de-Broglie wavelength, \(\lambda_e=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)
For photon of energy, \(E=h \nu=\frac{h c}{\lambda_p}\)
\(
\begin{aligned}
& \Rightarrow \lambda_p=\frac{h c}{E} \\
& \therefore \frac{\lambda_e}{\lambda_{p}}=\frac{h}{\sqrt{2 m E}} \times \frac{E}{h c}=\frac{1}{c}\left(\frac{E}{2 m}\right)^{1 / 2}
\end{aligned}
\)

Hint

(c) According to Einstein’s photoelectric equations.
\(
{e V}=\frac{h c}{\lambda}-\frac{h c}{\lambda_0}
\)
\(\therefore\) As per question, \(e V=\frac{h c}{\lambda}-\frac{h c}{\lambda_0} \dots(i)\)
\(
\frac{e V}{4}=\frac{h c}{2 \lambda}-\frac{h c}{\lambda_0} \dots(ii)
\)
From equations (i) and (ii), we get
\(
\begin{aligned}
& \frac{h c}{2 \lambda}-\frac{h c}{4 \lambda}=\frac{h c}{\lambda_0}-\frac{h c}{4 \lambda_0} \\
\Rightarrow & \frac{h c}{4 \lambda}=\frac{3 h c}{4 \lambda_0} \text { or } \lambda_0=3 \lambda
\end{aligned}
\)

Hint

(a) Let \(\phi_0\) be the work function of the surface of the material. Then,
According to Einstein’s photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is
\(
K_{\max _1}=\frac{h c}{\lambda}-\phi_0
\)
and that in the second case is
\(
K_{\max _2}=\frac{h c}{\frac{\lambda}{2}}-\phi_0=\frac{2 h c}{\lambda}-\phi_0
\)
But \(K_{\max _2}=3 K_{\max _1}\) (given)
\(
\therefore \quad \frac{2 h c}{\lambda}-\phi_0=3\left(\frac{h c}{\lambda}-\phi_0\right)
\)
\(
\begin{aligned}
& \frac{2 h c}{\lambda}-\phi_0=\frac{3 h c}{\lambda}-3 \phi_0 \\
& 3 \phi_0-\phi_0=\frac{3 h c}{\lambda}-\frac{2 h c}{\lambda} \\
& 2 \phi_0=\frac{h c}{\lambda} \quad \text { or } \quad \phi_0=\frac{h c}{2 \lambda}
\end{aligned}
\)

Hint

(a) According to Einstein’s photoelectric equation, the maximum kinetic energy of the emitted electron is
\(
K_{\max }=\frac{h c}{\lambda}-\phi_0
\)
where \(\lambda\) is the wavelength of incident light and \(\phi_0\) is the work function.
Here, \(\lambda=500 \mathrm{~nm}, h c=1240 \mathrm{eV} \mathrm{nm}\) and \(\phi_0=2.28 \mathrm{eV}\)
\(
\begin{aligned}
\therefore \quad K_{\max } & =\frac{1240 \mathrm{eV} \mathrm{~nm}}{500 \mathrm{~nm}}-2.28 \mathrm{eV} \\
& =2.48 \mathrm{eV}-2.28 \mathrm{eV}=0.2 \mathrm{eV}
\end{aligned}
\)
The de Broglie wavelength of the emitted electron is
\(
\lambda_{\min }=\frac{h}{\sqrt{2 m K_{\max }}}
\)
where \(h\) is the Planck’s constant and \(m\) is the mass of the electron.
As \(h=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}, m=9 \times 10^{-31} \mathrm{~kg}\) and \(K_{\max }=0.2 \mathrm{eV}=0.2 \times 1.6 \times 10^{-19} \mathrm{~J}\)
\(
\begin{aligned}
\therefore \quad \lambda_{\min } & =\frac{6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\sqrt{2\left(9 \times 10^{-31} \mathrm{~kg}\right)\left(0.2 \times 1.6 \times 10^{-19} \mathrm{~J}\right)}} \\
& =\frac{6.6}{2.4} \times 10^{-9} \mathrm{~m}=2.8 \times 10^{-9} \mathrm{~m}
\end{aligned}
\)
So, \(\lambda \geq 2.8 \times 10^{-9} \mathrm{~m}\)

Hint

(b) The energy \(E\) of the radiation can be expressed in terms of its frequency \(f\) using the equation:
\(
E=h \cdot f
\)
where \(h\) is Planck’s constant.
The momentum \(p\) of a photon (which is a quantum of radiation) can be expressed as:
\(
p=\frac{E}{c}
\)
where \(c\) is the speed of light. This can also be derived from the relationship:
\(
p=\frac{h}{\lambda}
\)
where \(\lambda\) is the wavelength of the radiation.
When the radiation falls on the surface, it has an initial momentum \(p_i\) :
\(
p_i=\frac{E}{c}
\)
Reflection of Radiation:
Since the surface is perfectly reflecting, the radiation will reflect back with the same magnitude of momentum but in the opposite direction. Thus, the final momentum \(p_f\) after reflection is:
\(
p_f=-\frac{E}{c}
\)
Calculating Change in Momentum:
The change in momentum \(\Delta p\) (which is the momentum transferred to the surface) can be calculated as:
\(
\Delta p=p_f-p_i
\)
Substituting the values of \(p_f\) and \(p_i\) :
\(
\Delta p=-\frac{E}{c}-\frac{E}{c}=-\frac{2 E}{c}
\)
Final Result:
The magnitude of the momentum transferred to the surface is:
\(
|\Delta p|=\frac{2 E}{c}
\)
The momentum transferred to the perfectly reflecting surface is \(\frac{2 E}{c}\).

Hint

(d) The Kinetic energy of charged particle accelerated under an applied potential is given by
\(
\Rightarrow q V=\frac{h c}{\lambda}-\phi
\)
The above equation can be rewritten as
\(
\Rightarrow \frac{h C}{\lambda}=q V+\phi \dots(1)
\)
For the first case, for \(\lambda\) and \(3 \mathrm{~V}_0\) the above equation can be written as
\(
\Rightarrow \frac{h c}{\lambda}=\phi+e 3 V_0 \dots(2)
\)
For the second case, for \(2 \lambda\) and \(V_0\) equation 1 can be written as
\(
\Rightarrow \frac{h c}{2 \lambda}=\phi+e V_0 \dots(3)
\)
Equation 3 – Equation 2 gives
\(
\begin{aligned}
& \Rightarrow \frac{h c}{\lambda}-\frac{h c}{2 \lambda}=\phi-\phi+3 e V_0-e V_0 \\
& \Rightarrow \frac{h c}{2 \lambda}=2 e V_0 \\
& \Rightarrow e V_0=\frac{h c}{4 \lambda}
\end{aligned}
\)
Substituting the value \(\phi=\frac{h c}{\lambda_0}\) ande \(V_0=\frac{h c}{4 \lambda}\) in equation 3 it can be rewritten as ( \(\lambda_0=\) Theshold wavelength)
\(
\begin{aligned}
& \Rightarrow \frac{h c}{2 \lambda}=\frac{h c}{\lambda_0}+\frac{h c}{4 \lambda} \\
& \Rightarrow \frac{h c}{2 \lambda}-\frac{h c}{4 \lambda}=\frac{h c}{4 \lambda}=\frac{h c}{\lambda_0}
\end{aligned}
\)
From the above equation \(\lambda_0=4 \lambda\)

Hint

(b) De-Brogle wavelength, \(\lambda=\frac{h}{p}\) or \(\lambda \propto \frac{1}{p}, \lambda p=\) constant
This represents a rectangular hyperbola.

Hint

(b) According to Einstein’s photoelectric equation,
The kinetic energy of emitted photoelectrons is
\(
K=h f-\phi_0
\)
where \(h f\) is the energy of incident radiation and \(\phi_0\) is work function of the metal.
As per question,
\(
\begin{aligned}
& 0.5 \mathrm{eV}=h f-\phi_0 \dots(i) \\
& 0.8 \mathrm{eV}=1.2 h f-\phi_0 \dots(ii)
\end{aligned}
\)
On solving eqns. (i) and (ii), we get
\(
\phi_0=1.0 \mathrm{eV}
\)

Hint

(b) de Broglie wavelength,
\(
\lambda=\frac{h}{\sqrt{2 m K}}
\)
where \(m\) is the mass and \(K\) is the kinetic energy of the particle.
When kinetic energy of the particle is increased to 16 times, then its de Broglie wavelength becomes,
\(
\lambda^{\prime}=\frac{h}{\sqrt{2 m(16 K)}}=\frac{1}{4} \frac{\lambda}{\sqrt{2 m K}}=\frac{\lambda}{4}(\operatorname{Using}(\mathrm{i}))
\)
\(\%\) change in the de Broglie wavelength
\(
\begin{aligned}
& =\frac{\lambda-\lambda^{\prime}}{\lambda} \times 100=\left(1-\frac{\lambda^{\prime}}{\lambda}\right) \times 100 \\
& =\left(1-\frac{1}{4}\right) \times 100=75 \%
\end{aligned}
\)

Hint

(b) Work function, \(\phi=h \nu\)
According to Einstein’s photoelectric equation
\(
\begin{aligned}
& \frac{1}{2} m v_{\max }^2=h(2 \nu)-h \nu \\
& \frac{1}{2} m v_{\max }^2=h \nu \\
& v_{\max }^2=\frac{2 h \nu}{m} \Rightarrow v_{\max }=\sqrt{\frac{2 h \nu}{m}}
\end{aligned}
\)

Hint

(d) Wavelength of an electron of energy \(E\) is
\(
\lambda_e=\frac{h}{\sqrt{2 m_e E}} \dots(i)
\)
Wavelength of a photon of same energy \(E\) is
\(
\lambda_p=\frac{h c}{E} \quad \text { or } E=\frac{h c}{\lambda_p} \dots(ii)
\)
Squaring both sides of Eq. (i), we get
\(
\lambda_e^2=\frac{h^2}{2 m_e E} \text { or } E=\frac{h^2}{2 m_e \lambda_e^2} \dots(iii)
\)
Equating (ii) and (iii), we get
\(
\begin{aligned}
& \frac{h c}{\lambda_p}=\frac{h^2}{2 m_e \lambda_e^2} \text { or } \lambda_p=\frac{2 m_e c}{h} \lambda_e^2 \\
& \lambda_p \propto \lambda_e^2
\end{aligned}
\)

Hint

(b) Given the initial distance between the light source and photocell: \(d_1=50 \mathrm{~cm}\)
Initial stopping potential: \(\boldsymbol{V}_{\mathbf{0}}\)
New distance between light source and photocell: \(d_2=25 \mathrm{~cm}\)

Stopping potential depends on the frequency of light, not the intensity.
Intensity of light is inversely proportional to the square of the distance from the source: \(I \propto \frac{1}{d^2}\).

How to solve?
Determine the new stopping potential by understanding that it is independent of the distance between the light source and the photocell.

Step 1: Understand the relationship between stopping potential and light intensity.
The stopping potential is determined by the maximum kinetic energy of the emitted photoelectrons.
The maximum kinetic energy of photoelectrons depends on the frequency of the incident light, not its intensity.
Changing the distance changes the intensity, but not the frequency.

Step 2: Apply the concept to the problem.
Since the frequency of light remains the same, the maximum kinetic energy of the photoelectrons remains the same.
Therefore, the stopping potential remains the same.

The new stopping potential will be \(V_0\).

Hint

(d) de Broglie wavelength of neutrons in thermal equilibrium at temperature \(T\) is
\(
\lambda=\frac{h}{\sqrt{2 m k_B T}}
\)
where \(m\) is the mass of the neutron
Here, \(m=1.67 \times 10^{-27} \mathrm{~kg}\)
\(
\begin{aligned}
& k_B=1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\
& h=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}
\end{aligned}
\)
\(
\therefore \quad \lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times T}}
\)
\(
\begin{aligned}
& =\frac{3.08 \times 10^{-34} \times 10^{25}}{\sqrt{T}} \mathrm{~m} \\
& =\frac{30.8 \times 10^{-10}}{\sqrt{T}}=\frac{30.8}{\sqrt{T}} Å
\end{aligned}
\)

Hint

(a) Energy of a photon,
\(=\frac{\left(6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{0.6 \times 10^{-6} \mathrm{~m}}=33 \times 10^{-20} \mathrm{~J}\)
Number of photons emitted per second is
\(
N=\frac{\frac{25}{100} P}{E}=\frac{\frac{25}{100} \times 200 \mathrm{~W}}{33 \times 10^{-20} \mathrm{~J}}=1.5 \times 10^{20}
\)

Hint

(d) Radius of the circular path of a charged particle in a magnetic field is given by
\(
\begin{aligned}
&R=\frac{m v}{B q} \text { or } m v=R B q\\
&\text { Here, } R=0.83 \mathrm{~cm}=0.83 \times 10^{-2} \mathrm{~m}
\end{aligned}
\)
\(
\begin{aligned}
B & =0.25 \mathrm{~Wb} \mathrm{~m}^{-2} \\
q & =2 e=2 \times 1.6 \times 10^{-19} \mathrm{C} \\
m v & =\left(0.83 \times 10^{-2}\right)(0.25)\left(2 \times 1.6 \times 10^{-19}\right)
\end{aligned}
\)
\(
\begin{aligned}
&\text { de Broglie wavelength, }\\
&\begin{aligned}
\lambda & =\frac{h}{m v}=\frac{6.6 \times 10^{-34}}{0.83 \times 10^{-2} \times 0.25 \times 2 \times 1.6 \times 10^{-19}} \\
& =0.01 Å
\end{aligned}
\end{aligned}
\)

Hint

(d) The de-Broglie wavelength associated with an electron is,
\(
\lambda=\frac{h}{P},
\)
or, \(P=\frac{h}{\lambda}\),
\(
\begin{aligned}
& \therefore \frac{\Delta P}{P}=-\frac{\Delta \lambda}{\lambda} \\
& \Rightarrow \frac{P}{P_{\text {initial }}}=\frac{0.5}{100}
\end{aligned}
\)
Therefore,
\(
P_{\text {initial }}=200 P
\)

Hint

(a) According to Einstein’s photoelectric equation
\(
\frac{1}{2} m v_{\max }^2=h \nu-\phi_0
\)
where \(\frac{1}{2} m v_{\max }^2\) is the maximum kinetic energy of the emitted electrons, \(h \nu\) is the incident energy and \(\phi_0\) is the work function of the metal.
\(
\therefore \quad \frac{1}{2} m v_{\max _1}^2=1 \mathrm{eV}-0.5 \mathrm{eV}=0.5 \mathrm{eV} \dots(i)
\)
and \(\frac{1}{2} m v_{\max _2}^2=2.5 \mathrm{eV}-0.5 \mathrm{eV}=2 \mathrm{eV} \dots(ii)\)
Divide (i) and (ii), we get
\(
\begin{aligned}
& \frac{v_{\max _1}^2}{v_{\max _2}^2}=\frac{0.5}{2} \\
& \frac{v_{\max _1}}{v_{\max _2}}=\sqrt{\frac{0.5}{2}}=\frac{1}{2}
\end{aligned}
\)

Hint

(c) For hydrogen atom, \(E_n=-\frac{13.6}{n^2} \mathrm{eV}\) For ground state, \(n=1\)
\(
\therefore \quad E_1=-\frac{13.6}{1^2}=-13.6 \mathrm{eV}
\)
For first excited state, \(n=2\)
\(
\therefore \quad E_2=-\frac{13.6}{2^2}=-3.4 \mathrm{eV}
\)
The energy of the emitted photon when an electron jumps from first excited state to ground state is
\(
h \nu=E_2-E_1=-3.4 \mathrm{eV_s}-(-13.6 \mathrm{eV})=10.2 \mathrm{eV}
\)
Maximum kinetic energy,
\(
K_{\max }={e V_s}=e \times 3.57 \mathrm{~V}=3.57 \mathrm{eV}
\)
According to Einstein’s photoelectric equation
\(
K_{\max }=h \nu-\phi_0
\)
where \(\phi_0\) is the work function and \(h \nu\) is the incident energy
\(
\phi_0=h \nu-K_{\max }=10.2 \mathrm{eV}-3.57 \mathrm{eV}=6.63 \mathrm{eV}
\)
Threshold frequency, \(\nu_0=\frac{\phi_0}{h}=\frac{6.63 \times 1.6 \times 10^{-19} \mathrm{~J}}{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}\)
\(
=1.6 \times 10^{15} \mathrm{~Hz}
\)

Hint

(a) When the nucleus emits a photon, the total momentum before and after the emission must be conserved. Initially, both the nucleus and the photon are at rest, so the total initial momentum is zero.
The momentum \(P\) of a photon can be expressed using the formula:
\(
P=\frac{E}{c}
\)
where \(E\) is the energy of the photon and \(c\) is the speed of light. The energy of the photon can be expressed in terms of its frequency \(\nu\) as:
\(
E=h \nu
\)
where \(h\) is Planck’s constant. Therefore, the momentum of the photon becomes:
\(
P_{\mathrm{photon}}=\frac{h \nu}{c}
\)
Let \(P_{\text {nucleus }}\) be the momentum of the recoiling nucleus. According to the conservation of momentum:
\(
P_{\text {nucleus }}+P_{\text {photon }}=0
\)
This implies:
\(
P_{\text {nucleus }}=-P_{\text {photon }}=-\frac{h \nu}{c}
\)
Relate Momentum to Recoil Energy
The momentum of the recoiling nucleus can also be expressed in terms of its mass \(M\) and its velocity \(v_r\) :
\(
P_{\text {nucleus }}=M v_r
\)
Setting the two expressions for momentum equal gives:
\(
M v_r=\frac{h \nu}{c}
\)
From this, we can solve for the recoil velocity \(v_r\) :
\(
v_r=\frac{h \nu}{M c}
\)
Calculate the Recoil Energy
The kinetic energy \(E_r\) of the recoiling nucleus can be expressed as:
\(
E_r=\frac{1}{2} M v_r^2
\)
Substituting the expression for \(v_r\) :
\(
E_r=\frac{1}{2} M\left(\frac{h \nu}{M c}\right)^2
\)
This simplifies to:
\(
E_r=\frac{1}{2} M \cdot \frac{h^2 \nu^2}{M^2 c^2}=\frac{h^2 \nu^2}{2 M c^2}
\)
Thus, the recoil energy of the nucleus when it emits a photon of frequency \(v\) is:
\(
E_r=\frac{h^2 \nu^2}{2 M c^2}
\)

Hint

(c) The stopping potential \(V_{\mathrm{s}}\) is relăted to the maximum kinetic energy of the emitted electrons \(K_{\max }\) through the relation
\(
\begin{aligned}
& K_{\max }=e V_s \\
& 0.5 \mathrm{eV}=e V_s \quad \text { or } \quad V_s=0.5 \mathrm{~V}
\end{aligned}
\)

Hint

(a) Here, work function, \(\phi_0=0.5 \mathrm{eV}\)
According to Einstein’s photoelectric equation
Maximum kinetic energy of the emitted electrons
\(=\) Incident photon energy – Work function
\(\therefore \quad K_{\max _1}=1 \mathrm{eV}-0.5 \mathrm{eV}=0.5 \mathrm{eV} \dots(i)\)
and \(K_{\max _2}=2.5 \mathrm{eV}-0.5 \mathrm{eV}=2 \mathrm{eV} \dots(ii)\)
Divide (i) by (ii), we get
\(
\begin{aligned}
& \frac{K_{\max _1}}{K_{\max _2}}=\frac{0.5 \mathrm{eV}}{2 \mathrm{eV}}=\frac{1}{4} \\
& \frac{\frac{1}{2} m v_{\max _1}^2}{\frac{1}{2} m v_{\max _2}^2}=\frac{1}{4} \text { or } \frac{v_{\max _1}}{v_{\max _2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}
\end{aligned}
\)

Hint

(a) The de Broglie wavelength \(\lambda\) associated with the electrons is
\(
\lambda=\frac{1.227}{\sqrt{V}} \mathrm{~nm}
\)
where \(V\) is the accelerating potential in volts.
or \(\quad \lambda \propto \frac{1}{\sqrt{V}}\)
\(\therefore \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{V_2}{V_1}}=\sqrt{\frac{100 \times 10^3}{25 \times 10^3}}=2\)
or \(\lambda_2=\frac{\lambda_1}{2}\)

Hint

(b) According to Einstein’s photoelectric equation
\(
e V_0=h f-h f_0
\)

where, \(f=\) Incident frequency
\(f_0=\) Threshold frequency
\(V_0=\) Cut-off or stopping potential
or \(\quad V_0=\frac{h}{e}\left(f-f_0\right)\)
Substituting the given values, we get
\(
V_0=\frac{6.63 \times 10^{-34}\left(8.2 \times 10^{14}-3.3 \times 10^{14}\right)}{1.6 \times 10^{-19}} \approx 2 \mathrm{~V}
\)

Hint

(d) According to Einstein’s photoelectric equation
\(
K_{\max }=h f-h f_0
\)
Since \(K_{\max }\) is +ve , the photoelectric emission occurs only if
\(
h f>h f_0 \text { or } f>f_0
\)
The photoelectric emission occurs only when the incident light has more than a certain minimum frequency. This minimum frequency is called threshold frequency.

Hint

(a) increasing the potential difference between the anode and filament.

Explanation:
In the Davisson and Germer experiment, the velocity of electrons is directly proportional to the potential difference applied between the anode and the filament. Increasing the potential difference creates a stronger electric field, accelerating the electrons to a higher velocity.

Why other options are incorrect:
(b) Increasing the filament current:

The filament current controls the heating of the filament, which influences the number of electrons emitted, not their velocity. While a higher filament current might produce more electrons, their velocity depends on the potential difference, not the emission rate.
(c) Decreasing the filament current:

Decreasing the filament current would reduce the number of electrons emitted, but not their velocity.
(d) Decreasing the potential difference between the anode and filament:

Decreasing the potential difference would decrease the accelerating force on the electrons, resulting in a lower velocity.

Hint

(a) For a source \(S_1\),
Wavelength \(\lambda_1 \equiv 5000 Å\)
Number of photons emitted per second, \(N_1=10^{15}\)
Energy ot each photon, \(E_1=\frac{h c}{\lambda_1}\)
Power of source \(S_1, P_1=E_1 N_1=\frac{N_1 h c}{\lambda_1}\)
For a source \(S_2\),
Wavelength, \(\lambda_2=5100 Å\)
Number of photons emitted per second,
\(
N_2=1.02 \times 10^{15}
\)
Energy of each photon, \(E_2=\frac{h c}{\lambda_2}\)
Power of source \(S_2, P_2=N_2 E_2=\frac{N_2 h c}{\lambda_2}\)
\(
\begin{aligned}
& \therefore \frac{\text { Power of } S_2}{\text { Power of } S_1}=\frac{P_2}{P_1}=\frac{\frac{N_2 h c}{\lambda_2}}{\frac{N_1 h c}{\lambda_1}}=\frac{N_2 \lambda_1}{N_1 \lambda_2} \\
& \quad=\frac{\left(1.02 \times 10^{15} \text { photons } / \mathrm{s}\right) \times(5000 Å)}{\left(10^{15} \text { photons } / \mathrm{s}\right) \times(5100 Å)}=\frac{51}{51}=1
\end{aligned}
\)

Hint

(b) Here,
Incident wavelength, \(\lambda=200 \mathrm{~nm}\)
Work function, \(\phi_0=5.01 \mathrm{eV}\)
According to Einstein’s photoelectric equation
\(
\begin{aligned}
& e V_{\mathrm{s}}=h \nu-\phi_0 \\
& e V_{\mathrm{s}}=\frac{h c}{\lambda}-\phi_0
\end{aligned}
\)
where \(V_s\) is the stopping potentia
\(
\begin{aligned}
e V_s & =\frac{(1240 \mathrm{eV} \mathrm{~nm})}{(200 \mathrm{~nm})}-5.01 \mathrm{eV} \\
& =6.2 \mathrm{eV}-5.01 \mathrm{eV}=1.2 \mathrm{eV}
\end{aligned}
\)
Stopping potential, \(V_s=1.2 \mathrm{~V}\)
The potential difference that must be applied to stop photoelectrons \(=-V_s=-1.2 \mathrm{~V}\)

Hint

(a) Intensity of radiation: \({I}\)
Number of photoelectrons: \({N}\)
Maximum kinetic energy: \({T}\)
New intensity of radiation: \({2 I}\)

The number of photoelectrons is directly proportional to the intensity of the incident radiation.
The maximum kinetic energy of photoelectrons depends on the frequency of the incident radiation, not the intensity.

How to solve?
Determine the new number of photoelectrons and their maximum kinetic energy when the intensity is doubled.

Step 1: Determine the new number of photoelectrons
The number of photoelectrons is directly proportional to the intensity. If the intensity is doubled, the number of photoelectrons is also doubled.
New number of photoelectrons: \(2 N\)

Step 2: Determine the new maximum kinetic energy
The maximum kinetic energy depends on the frequency, not the intensity. If the intensity changes, the maximum kinetic energy remains the same.

New maximum kinetic energy: {T}[/latex]
The number of emitted electrons is \(2 N\) and their maximum kinetic energy is \(T\).

Alternate Explanation:

Photons: Light is said to be made up of small packets known as photons. The energy of each photon is given as
\(
E=h f
\)
\(f\) is the frequency of lightwave, \(h\) is Planck’s constant.
Photoelectrons: The electrons emitted from the metal when a ray of light incident on it is known as a photoelectron. This phenomenon is known as the photoelectric effect.
The maximum kinetic energy of electron depends upon the energy of photons falling on the metal surface.
\(
K=h f-\phi
\)
where \(\phi\) is the minimum energy required to emit an electron out of metal.
More Photons will make more photoelectron to eject but the energy of photoelectron will depend upon the energy of the incident photon and not on the number of photons.
Intensity: Intensity is defined as the number of photons per unit area. It depends upon the number of photons.
\(
{I}={n} / {A}
\)
\({n}=\) number of the photon, \({A}=\) Area where the photon is falling
If the area is fixed, \(I \propto {n}\)

Hint

(d) Energy released when electron in the atom jumps from excited state \((n=3)\) to ground state \((n=1)\) is
\(
\begin{aligned}
& E=h f=E_3-E_1=\frac{-13.6}{3^2}-\left(\frac{-13.6}{1^2}\right) \\
& =\frac{-13.6}{9}+13.6=12.1 \mathrm{eV}
\end{aligned}
\)
Therefore, stopping potential
\(
\begin{aligned}
& e V_0=h f-\phi_0 \\
& \quad=12.1-5.1 \quad\left[\because \text { work function } \phi_0=5.1\right] \\
& V_0=7 \mathrm{~V}
\end{aligned}
\)

Hint

(b) \(\lambda=6670 Å\)
\(
E \text { of a photon }=\frac{12400 \mathrm{eV Å}}{6670 Å}=\frac{12400}{6670} \times 1.6 \times 10^{-19} \mathrm{~J} .
\)
Energy emitted per second, power \(P=9 \times 10^{-3} \mathrm{~J}\)
\(\therefore \quad\) Number of photons incident \(=\frac{\text { Power }}{\text { Energy }}=\frac{P}{E}\)
\(=\frac{9 \times 10^{-3} \times 6670}{12400 \times 1.6 \times 10^{-19}}=3 \times 10^{16}\).

Hint

(b)

The energy of the incident photon is \({E =} {h f}\), hence frequency determines the energy of incident photons.

This energy is imparted on the photosensitive surface to cause photo ejection.
At the points where the curve cuts the \(x\)-axis, the incident photon energy is equal to the work potential of the material. Hence energy, hence frequency, is the same for radiation of curves \({a} \& {b}\).

Intensity is the measure of the number of photons striking per unit of time.
Thus they determine the amount of photocurrent produced thus it is the same for \({b}\) & c.

Hint

(c) The number of photoelectrons emitted is directly proportional to the intensity of the light, not the frequency or threshold frequency.

Hint

(a) Work function: \(\phi=6.2 \mathrm{eV}\)
Stopping potential: \(V_0=5 \mathrm{~V}\)
Energy of a photon: \(E=h f=\frac{h c}{\lambda}\)
Photoelectric effect equation: \(E=\phi+e V_0\) \(h c \approx 1240 \mathrm{eV} \mathrm{nm}\)
Use the photoelectric effect equation: \(E=\phi+e V_0\).
Substitute the given values: \(E=6.2 \mathrm{eV}+e(5 \mathrm{~V})=6.2 \mathrm{eV}+5 \mathrm{eV}\).
Calculate the result: \({E}=11.2 \mathrm{eV}\).
Use the formula: \(\lambda=\frac{h c}{E}\).
Substitute the values: \(\lambda=\frac{1240 \mathrm{eV} \cdot \mathrm{nm}}{11.2 \mathrm{eV}}\).
Calculate the result: \(\lambda \approx 110.7 \mathrm{~nm}\).
The wavelength \(\lambda \approx 110.7 \mathrm{~nm}\) lies in the ultraviolet region.
The wavelength of the incident radiation lies in the ultraviolet region.

Hint

(d) When a sufficient voltage is applied in a discharge tube, the gas dissipates and becomes conducting. In the discharge tube collision between charged particles(electrons) emitted from the cathode and atoms of the gas takes place and forms cathode rays which on striking on the walls give a fluorescence effect. This gives a colored glow to the tube.

Hint

(a) 
\(
\begin{aligned}
\lambda & =\frac{h}{10^{-6} \mathrm{~kg} \times v} \\
& =\frac{h}{9.1 \times 10^{-31} \mathrm{~kg} \times 3 \times 10^6 \mathrm{~m} / \mathrm{s}}
\end{aligned}
\)
\(
\therefore \quad v=2.7 \times 10^{-18} \mathrm{~m} / \mathrm{s} .
\)

Hint

(a) For a light source of power \(P\) watt, the intensity at a distance \(d\) is given by
\(
I=\frac{P}{4 \pi d^2}
\)
where we assume light to spread out uniformly in all directions i.e. it is a spherical source.
\(\therefore \quad I \propto \frac{1}{d^2} \quad\) or \(\quad \frac{I_1}{I_2}=\frac{d_2^2}{d_1^2}\)
or, \(\frac{I_1}{I_2}=\left(\frac{1}{0.5}\right)^2 \quad\) or, \(\quad \frac{I_1}{I_2}=4 \quad\) or, \(\quad I_2=\frac{I_1}{4}\).
In a photoelectric emission, the number of photoelectrons liberated per second from a photosensitive metallic surface is proportional to the intensity of the light. When a intensity of source is reduced by a factor of four, the number of photoelectrons is also reduced by a factor of 4.

Hint

(a) Power of monochromatic light beam is \(P=N h f\) where \(N\) is the number of photons emitted per second.
Power \(P=2 \times 10^{-3} W\)
Energy of one photon \(E=h f\)
\(
=6.63 \times 10^{-34} \times 6 \times 10^{14} J
\)
Number of photons emitted per second,
\(
\begin{aligned}
& N=P / E \\
& =\frac{2 \times 10^{-3}}{6.63 \times 10^{-34} \times 6 \times 10^{14}} \\
& =0.05 \times 10^{17}=5 \times 10^{15}
\end{aligned}
\)

Hint

(a) When the electric field is switched off, and only the magnetic field remains, the electrons will move in a circular orbit (A) because the magnetic force will act as a centripetal force, causing them to move in a circular path.

Here’s a more detailed explanation:
Initial State:
The electrons initially pass through both electric and magnetic fields without deflection, meaning the electric force and magnetic force are equal and opposite, resulting in a net force of zero.

Switching off Electric Field:
When the electric field is switched off, the only force acting on the electrons is the magnetic force.

Magnetic Force and Circular Motion:
The magnetic force is always perpendicular to both the velocity of the electron and the magnetic field, causing the electron to move in a circular path.

Centripetal Force:
The magnetic force provides the necessary centripetal force to maintain the circular motion.

Hint

(c)  The photoelectric current is directly proportional to the intensity of illumination. Therefore a change in the intensity of the incident radiation will change the photocurrent also.

Hint

(d) Let \(K\) and \(K^{\prime}\) be the maximum kinetic energy of photoelectrons for incident light of frequency \(\nu\) and \(2 \nu\) respectively.
According to Einstein’s photoelectric equation,
\(\)
\begin{aligned}
K & =h \nu-E_0 \dots(i) \\
\text { and } K^{\prime} & =h(2 \nu)-E_0 \dots(ii) \\
& =2 h \nu-E_0=h \nu+h \nu-E_0 \\
K^{\prime} & =h \nu+K \quad[\text { using (i) }]
\end{aligned}
\(\)

Hint

(d) Energy of photon \(E=1 \mathrm{MeV}\) Momentum of photon \(p=E / c\)
\(
\begin{aligned}
\therefore p=\frac{E}{c}= & \frac{1 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J}}{3 \times 10^8 \mathrm{~ms}^{-1}}=0.53 \times 10^{-21} \\
& \approx 5 \times 10^{-22} \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(c)

In a discharge tube, ionization of enclosed gas is produced due to collisions between (c) negative electrons and neutral atoms/molecules.

Explanation:
In a discharge tube, a high voltage creates an electric field that accelerates electrons to high speeds. These fast-moving electrons then collide with neutral gas atoms or molecules, knocking off electrons and creating positively charged ions. This process, called ionization, continues as the freed electrons collide with other neutral atoms/molecules, further ionizing the gas.

Why other options are incorrect:
(a) neutral gas atoms/molecules:

Neutral atoms/molecules don’t initiate ionization; they are the target of the ionizing electrons.
(b) positive ions and neutral atoms/molecules:

Positive ions are already charged and don’t typically cause ionization through collisions with neutral atoms/molecules. They are attracted to the cathode.
(d) photons and neutral atoms/molecules:

While photons can interact with atoms/molecules, the energy required for ionization in a discharge tube is usually much higher than that of typical photons. In most discharge tubes, the ionization process is primarily due to electron collisions.

Hint

(d) The solution to our problem consists, in Einsteins photoelectric equation.
Einsteins photoelectric equation can be written as
\(
\begin{aligned}
& \frac{1}{2} m v^2=h \nu-\phi \\
& \Rightarrow \frac{1}{2} m \times\left(4 \times 10^6\right)^2=2 h \nu_0-h \nu_0 \dots(i)
\end{aligned}
\)
and \(\frac{1}{2} m \times v^2=5 h \nu_0-h \nu_0 \quad \ldots(ii)\)
Dividing Eq. (ii) by (i), we get
\(
\begin{aligned}
& \frac{v^2}{\left(4 \times 10^6\right)^2}=\frac{4 h \nu_0}{h \nu_0} \\
& \Rightarrow v^2=4 \times 16 \times 10^{12} \\
& \Rightarrow v^2=64 \times 10^{12} \\
& \therefore v=8 \times 10^6 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& \text { Energy of the incident radiation }=\frac{h c}{\lambda} \\
& =\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{41 \times 10^{-8}} \mathrm{~J} \\
& =\frac{19.89 \times 10^{-18}}{41 \times 1.6 \times 10^{-19}} \mathrm{eV}=3 \mathrm{eV}
\end{aligned}
\)
Formetals A and \(\mathrm{B}, W_A=1.92 \mathrm{eV}\) and \(W_B=2 \mathrm{eV}\)
\(\therefore\) This is less than 3 eV , the energy of incident radiation.
\(\therefore\) Photoemission is possible for \(A\) and \(B\).
However for C, \(W_C=5 \mathrm{eV}\). Hence the photoemission is not possible.

Hint

(c) The maximum kinetic energy of photoelectron ejected is given by
\(
\mathrm{K} . \mathrm{E} .=h f-W=h f-h f_0
\)
where work function depends on the type of material. If the frequency of incident radiation is greater than \(f_0\) only then the ejection of photoelectrons start. After that as frequency inereases kinetic energy also increases

Hint

(c) \(J J\) Thomson’s cathode-ray tube experiment demonstrated the relation for \(e / m\) of charged particles. The relation is
\(
\frac{e}{m}=\frac{E^2}{2 B^2 V}
\)
Thus, knowing \(E, B\) and \(V\), the value of \(e / m\) for electrons and for protons can be calculated. It is found the \(e / m\) of electrons is much greater than the \(e / m\) of protons.
\(\left(\frac{e}{m}\right)_{\text {proton }}=\frac{1}{1837}\left(\frac{e}{m}\right)_{\text {electron }}\)
\(\therefore\left(\frac{e}{m}\right)_{\text {proton }} \ll\left(\frac{e}{m}\right)_{\text {electron }}\)

Hint

(d) Photoelectric current \(I \propto\) intensity of light and intensity \(\propto \frac{1}{(\text { distance })^2}\)
\(
\therefore I \propto \frac{1}{(\text { distance })^2} .
\)

Note: When distance is doubled, intensity becomes one-fourth of initial value.

Hint

(b) Cathode rays are basically negatively charged particles (electrons). If the cathode rays are allowed to pass between two plates kept at a difference of potential, the rays are found to be deflected from the rectilinear path. The direction of deflection shows that the rays carry negative charges.

Hint

(b) Alkali metals, viz, lithium, sodium, potassium shows photoelectric effect even with visible light. Photo-electric emission depends on the work function ( \(W\) ) of the metal surface. If the frequency \(\nu\) of the incident ray is less than certain frequency \(\nu_0\) known as threshold frequency, no photoelectric effect is possible. \(W=h \nu_0\), is different for different materials.

Therefore, when ultra-violet rays fail to emit photoelectrons, rays like X-rays with higher frequency should be used.

Explanation: The photoelectric effect requires light with a frequency above the metal’s threshold frequency to eject electrons from the surface. X-rays have a much higher frequency than ultraviolet light, so they are more likely to trigger the photoelectric effect.
Why other options are incorrect:
a. Infrared rays:
Infrared rays have a lower frequency than visible light, and even lower than ultraviolet light, so they won’t have enough energy to cause the photoelectric effect.
c. Radio wave:
Radio waves have an even lower frequency than infrared rays, making them incapable of triggering the photoelectric effect.
d. Lightwave:
While the question states the photoelectric effect doesn’t occur with ultraviolet rays, “lightwave” is too broad of a term. Light encompasses a wide range of frequencies, including ultraviolet, visible light, and infrared. Since the question specifies ultraviolet rays don’t work, “lightwave” doesn’t provide a correct answer.

Hint

(d) \(\lambda=\frac{{h}}{{mv}}\)
Where \(m, v\) are the mass and velocity. All the particles are moving with same velocity, the particle with least mass will have maximum de-Broglie wavelength. \(\beta\) particles has the lowest mass and therefore it has maximum wavelength.

Hint

(d) The value of Planck’s constant is \(6.63 \times 10^{-34} \mathrm{~J}-\) sec.

Hint

(b)

\(
I \propto \frac{1}{d^2}
\)
Let’s denote:
The initial distance from the light source to the photocell as \(d\).
The initial intensity of light at this distance as \(I_1\).
The new distance as \(\frac{d}{2}\).
The new intensity at this distance as \(I_2\).
\(
\begin{aligned}
& I_1 \propto \frac{1}{d^2} \\
& I_2 \propto \frac{1}{\left(\frac{d}{2}\right)^2}=\frac{1}{\frac{d^2}{4}}=\frac{4}{d^2}
\end{aligned}
\)
\(
\frac{I_2}{I_1}=\frac{\frac{4}{d^2}}{\frac{1}{d^2}}=4
\)

The number of electrons emitted per second from the photocell is directly proportional to the intensity of the light:
\(
N_1 \propto I_1 \quad \text { and } \quad N_2 \propto I_2
\)
Where \(N_1\) is the number of electrons emitted at distance \(d\) and \(N_2\) is the number emitted at distance \(\frac{d}{2}\).
From the intensity ratio, we can relate the number of emitted electrons:
\(
\frac{N_2}{N_1}=\frac{I_2}{I_1}=4
\)
Thus, we can express \(N_2\) in terms of \(N_1\) :
\(
N_2=4 N_1
\)
The number of electrons emitted per second will be 4 times the original number.

Hint

(a) The correct answer is (a) photoelectric effect.
Explanation:
The photoelectric effect demonstrates the particle nature of light because it shows that light can interact with electrons in a metal like particles, knocking them out of the metal. This phenomenon can’t be explained by wave theory alone.

Why other options are incorrect:
Interference:
Interference, diffraction, and polarization are all wave phenomena. They involve the superposition of waves and can explain phenomena like light bending around obstacles or the changing of light’s polarization. These phenomena can’t be explained by the particle nature of light.

Refraction:
Refraction is the bending of light as it passes from one medium to another. This is also a wave phenomenon. It can be explained by the change in the speed of light in different mediums.

Polarization:
Polarization is the process of aligning the vibrations of light waves in a specific direction. This is a wave phenomenon. It deals with the orientation of the electric field of light waves, not its particle nature.

Hint

(a) The force acting on a charge \(Q\) in an electric field \(\vec{E}\) is given by:
\(
F_E=Q \vec{E}
\)

The force acting on a charge \(Q\) moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is given by:
\(
F_B=Q(\vec{v} \times \vec{B})
\)

Set the Forces Equal:
For the electron beam to be unaffected, these two forces must be equal in magnitude:
\(
F_E=F_B
\)
This leads to the equation:
\(
Q \vec{E}=Q(\vec{v} \times \vec{B})
\)

Cancel the Charge:
Since the charge \(Q\) is the same on both sides, we can cancel it out (assuming
\(
\begin{aligned}
& Q \neq 0): \\
& \vec{E}=\vec{v} \times \vec{B}
\end{aligned}
\)

Magnitude of the Velocity:
Since \(\vec{E}\) and \(\vec{B}\) are perpendicular, we can express the magnitude of the velocity \(v\) as:
\(
E=v B
\)
Rearranging this gives:
\(
v=\frac{E}{B}
\)
Final Expression:
Therefore, the velocity of the underreflected electron beam is:
\(
v=\frac{E}{B}
\)
Conclusion: The velocity of the undeflected electron beam in the Thomson mass spectrograph when the electric field is perpendicular to the magnetic field is given by:
\(
v=\frac{E}{B}
\)

Hint

(a) Einstein work on photoelectric effect supports the equation \(E=h \nu\). It is based on quantum theory of light.

Explanation:
The photoelectric effect is the emission of electrons from a metal when light shines on it.
Einstein explained the photoelectric effect by proposing that light is quantized into packets of energy called photons.
The energy of a photon is given by \(E=h \nu\), where \(h\) is Planck’s constant and \(\nu\) is the frequency of the light.

Hint

(b) Miliikan evaluated the mass of electron indirectly with the help of charge by oil drop experiment.

Hint

(b) We need to find the kinetic energy of an electron given its total energy.
Given, total energy of the electron \(E_{\text {total }}=3.555 \mathrm{MeV}\) (known fact).
Rest mass energy of an electron \(E_0=0.51 \mathrm{MeV}\).

The total energy of a particle is the sum of its kinetic energy and rest mass energy.
The formula for total energy is \(E_{\text {total }}=K E+E_0\).

How to solve?
Subtract the rest mass energy of the electron from its total energy to find the kinetic energy.

Step 1: Calculate the kinetic energy
Use the formula \({K E}={E}_{\text {total }}-{E}_{{0}}\).
Substitute the given values:
\(K E=3.555 \mathrm{MeV}-0.51 \mathrm{MeV}\)
\(K E=3.045 \mathrm{MeV}\)

The kinetic energy of the electron is 3.045 MeV.

Hint

(c) The kinetic energy of the electron \(\left(K E_e\right)\) is equal to the kinetic energy of the photon ( \(K E_{p h}\) ).

De-Broglie wavelength for a particle is given by \(\lambda=\frac{{h}}{{p}}\), where \({h}\) is Planck’s constant and \(p\) is momentum.
Kinetic energy of an electron is \(K E_e=\frac{p_e^2}{2 m_e}\), where \(p_e\) is the momentum of the electron and \(m_e\) is the mass of the electron.
Kinetic energy of a photon is \(K E_{p h}=p c\), where \(p\) is the momentum of the photon and \(c\) is the speed of light.

How to solve?
Equate the kinetic energies of the electron and photon, then use the deBroglie wavelength formula to find the relationship between their wavelengths.

Step 1: Equate the kinetic energies of the electron and photon
\(
\begin{aligned}
K E_e & =K E_{p h} \\
\frac{p_e^2}{2 m_e} & =p_{p h} c
\end{aligned}
\)
Step 2: Express the momenta in terms of de-Broglie wavelengths
\(
\begin{aligned}
& p_e=\frac{h}{\lambda_e} \\
& p_{p h}=\frac{h}{\lambda_{p h}}
\end{aligned}
\)
Step 3: Substitute the momenta into the kinetic energy equation
\(
\begin{aligned}
& \frac{\left(\frac{h}{\lambda_e}\right)^2}{2 m_e}=\frac{h}{\lambda_{p h}} c \\
& \frac{h^2}{2 m_e \lambda_e^2}=\frac{h c}{\lambda_{p h}}
\end{aligned}
\)
Step 4: Simplify and solve for the relationship between \(\lambda_{p h}\) and \(\lambda_c\)
\(
\begin{aligned}
& \frac{h}{2 m_e \lambda_e^2}=\frac{c}{\lambda_{p h}} \\
& \lambda_{p h}=\frac{2 m_e c \lambda_e^2}{h} \\
& \lambda_{p h} \propto \lambda_e^2 \\
& \lambda_{p h}>\lambda_e
\end{aligned}
\)
The de-Broglie wavelength of the photon is greater than the de-Broglie wavelength of the electron.

Hint

(c)

\(
\begin{aligned}
& \frac{f c}{\lambda}=\phi+K \cdot E \cdot \\
& K \cdot E=\frac{h c}{\lambda}-\phi
\end{aligned}
\)
\(
\begin{aligned}
& K \cdot E_1=\frac{h c}{3000}-\phi \dots(1) \\
& K \cdot E_2=\frac{h c}{2000}-\phi \dots(2)
\end{aligned}
\)
\(
\begin{aligned}
&\text { Eqn (1) – Eqn (2) }\\
&K E_1-K E_2=\frac{h c}{3000}-\frac{h c}{2000} <0
\end{aligned}
\)
\(
K E_2>K E_1
\)

Hint

(b) Given, Amplitude of light: \({A}\)
Wavelength of light: \(\lambda\)
Cut-off wavelength: \({\lambda}_{{0}}\)

The intensity of light is proportional to the square of its amplitude: \(I \propto A^2\).
The photoelectric effect occurs only if the wavelength of the incident light is less than the cut-off wavelength: \(\lambda<\lambda_0\).
The saturation current is directly proportional to the intensity of the incident light: \(I_{\text {sat }} \propto I\).

How to solve?
Determine the relationship between saturation current and the amplitude of light when the photoelectric effect occurs.

Step 1: Check the condition for photoelectric effect
Photoelectric effect occurs only if \(\lambda<\lambda_0\).
If \(\lambda>\lambda_0\), no current flows.

Step 2: Relate intensity to amplitude
Intensity of light is proportional to the square of the amplitude:
\(\quad I \propto A^2\)

Step 3: Relate saturation current to intensity
Saturation current is proportional to the intensity of light:
\(I_{\text {sat }} \propto I\)

Step 4: Combine the relationships
Since \(I_{\text {sat }} \propto I\) and \(I \propto A^2\), then \(I_{\text {sat }} \propto A^2\).

The saturation current is proportional to the square of the amplitude of the incident light when the wavelength is less than the cut-off wavelength.

Hint

(d) In a discharge tube, the gas is ionized by a high voltage.
This ionization process creates positive ions and free electrons.
Both positive ions and electrons contribute to the current flow.
The current conduction in a discharge tube is due to positive ions and electrons.

Hint

(a) \(\mathrm{K} . \mathrm{E} .=1.6 \times 10^{-19} \times 1=1 \mathrm{eV}\)

Hint

(a)
\(
\begin{aligned}
& \phi=h \mathrm{\nu}=\frac{h c}{\lambda} \\
& \Rightarrow \lambda=\frac{h c}{\phi}=\frac{1242 \mathrm{eV} . \mathrm{nm}}{4.125} \approx 3000 Å
\end{aligned}
\)

Hint

(c) If the intensity of light of a given frequency is increased, then the number of photons striking the surface per second will increase in the same ratio. This increased number of photons strikes more electrons of metals and hence, the number of photolectrons emitted through the surface increases and hence photoelectric current increases.

Hint

(d) According to Einstein’s photoelectric equation,
\(
\frac{1}{2} m v^2=\frac{h c}{\lambda}-\phi_0 \quad \text { or, } \quad \frac{h c}{\lambda}=\frac{1}{2} m v^2+\phi_0
\)
and \(\frac{1}{2} m v_1^2=\frac{h c}{3 \lambda / 4}-\phi_0=\frac{4}{3}\left(\frac{1}{2} m v^2+\phi_0\right)-\phi_0\)
So, \(v_1\) is greater than \(v(4 / 3)^{1 / 2}\).

Hint

(b)

\(
\begin{aligned}
& K . E_{\max }=\frac{h c}{\lambda}-\phi \\
& =(12400 \mathrm{eV Å}) /(5000 Å)-1.5 \mathrm{eV} \\
& =(2.48-1.5) \mathrm{eV}=0.98 \mathrm{eV}
\end{aligned}
\)

Hint

(a) Since the emission of photoelectrons is directly proportional to the intensity of the incident light, therefore photocurrent increases with the intensity of light.

Hint

(c) Potential difference \((V)=100\) volts. Kinetic energy of an electron (K.E.)
\(
=e V=\left(1.6 \times 10^{-19}\right) \times 100=1.6 \times 10^{-17} \text { joule. }
\)

Hint

(c) Kinetic energy \((E)=100 \mathrm{eV}\);
\(
\begin{aligned}
& \text { Mass of electron }(m)=9.1 \times 10^{-31} \mathrm{~kg} \\
& \quad 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J} \text { and } \\
& \text { Planck’s constant }(h)=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s} \text {. } \\
& \text { Energy of an electron }(E)=100 \times\left(1.6 \times 10^{-19}\right) \mathrm{J}
\end{aligned}
\)
\(
\begin{aligned}
\lambda & =\frac{h}{\sqrt{2 m E}}=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 100 \times 1.6 \times 10^{-19}}} \\
& =1.2 \times 10^{-10} \mathrm{~m}=1.2 Å .
\end{aligned}
\)

Hint

(a) At 0.2 mm of Hg pressure in the discharge tube, Crooke’s dark space with glow near the electrodes occurs, while at 0.5 mm of Hg pressure in the discharge tube, Faraday’s dark space occurs.

Hint

(a) The kinetic energy of an electron \(\frac{1}{2} \times m v^2=e V\) final velocity of the electron \((v)=\sqrt{\frac{2 e V}{m}}\).

Hint

(c) Energy of the photon \(E=\frac{h c}{\lambda}\) or \(\lambda=\frac{h c}{E}\), where \(\lambda\) is the wavelength.

Hint

(c) The velocity of a photon is given by
\(
C=\nu \lambda
\)
From the above equation, it is clear that velocity is directly proportional to frequency (\(\nu\)).

Hint

(d) Momentum of electrons \(\left(p_{\mathrm{e}}\right)=\sqrt{2 m e V}\) and momentum for proton \(\left(p_p\right)=\sqrt{2 \mathrm{MeV}}\)
\(
\text { Therefore, } \frac{\lambda_{{p}}}{\lambda_{{e}}}=\frac{h / p_p}{h / p_e}=\frac{p_e}{p_p} \equiv \frac{\sqrt{2 m e V}}{\sqrt{2 M e V}}=\sqrt{\left(\frac{m}{M}\right)} \text {. }
\)
\(
\text { Therefore } \lambda_p=\lambda_e \sqrt{\left(\frac{m}{M}\right)}=\lambda \sqrt{\left(\frac{m}{M}\right)} \text { as } \lambda_e = \lambda
\)

Hint

(a) Wavelength \((\lambda)=\frac{h}{m v}=\frac{h}{p}\). Therefore for the same wavelength of electrons and photons, the momentum should be same.

Hint

(c) The correct answer is Electrons.
Explanation:
When a surface is heated, the thermal energy excites the electrons within the material. Some of these electrons gain enough energy to escape the surface, leaving the material behind with a positive charge. This process is called thermionic emission. The emitted particles are electrons.

Why other options are incorrect:
(a) Protons: Protons are part of the nucleus of an atom, and are held tightly together by the strong nuclear force. It requires immense energy to remove a proton from an atom, much more than what is provided by typical heating. Thermionic emission only involves the ejection of loosely bound electrons.
(b) Neutrons: Like protons, neutrons are also located in the nucleus. They are neutral particles, meaning they have no electric charge. Since thermionic emission involves the ejection of charged particles (electrons), neutrons cannot be knocked out by heat.
(d) Nuclei: Nuclei are composed of protons and neutrons. As explained above, these particles are held very tightly within the nucleus and are not ejected during thermionic emission. The process only concerns the escape of loosely bound electrons from the surrounding electron cloud.

Hint

\(
\begin{aligned}
& \text { (b) Threshold wavelength }(\lambda)=2000 Å \\
& =2000 \times 10^{-10} \mathrm{~m} \text {. } \\
& \text { Work function }(W)=\frac{h c}{\lambda} \\
& =\frac{\left(6.6 \times 10^{34}\right) \times\left(3 \times 10^8\right)}{2000 \times 10^{-10}} \\
& =9.9 \times 10^{-19} \mathrm{~J}=\frac{9.9 \times 10^{-19}}{1.6 \times 10^{-19}}=6.2 \mathrm{eV}
\end{aligned}
\)

Hint

(a) Potential difference \((\mathrm{V})=100 \mathrm{~V}\). The kinetic energy of an electron \(=1 \mathrm{eV}=1 \times\left(1.6 \times 10^{-19}\right)\) \(=1.6 \times 10^{-19} \mathrm{~J}\). Therefore kinetic energy in 100 volts \(=\left(1.6 \times 10^{-19}\right) \times 100=1.6 \times 10^{-17} \mathrm{~J}\).

Hint

(b) Gases begin to conduct electricity at low pressure because at low pressure, colliding electrons can acquire higher kinetic energy due to increased mean free path leading to ionization of atoms.

Explanation:
At normal pressure, gas molecules are close together, frequently colliding with each other. This prevents electrons from moving freely and conducting electricity. However, when the pressure is lowered, the gas molecules are further apart, meaning the electrons have a greater chance of traveling a longer distance before colliding with another molecule. This increased “mean free path” allows electrons to gain more kinetic energy, which can then ionize other atoms, creating free electrons and positive ions. This process continues, forming a chain reaction and enabling the gas to conduct electricity.

Why other options are incorrect:
(a) At low pressures gases turn of plasma:

While plasma is a state of matter where electrons are free from their atoms, it’s not the direct reason why gases conduct electricity at low pressure. Plasma formation is a result of high energy input into a gas, such as high temperature or strong electric fields. Low pressure simply facilitates the ionization process that leads to plasma formation under certain conditions, but it’s not the cause of conductivity itself.
(c) Atoms break up into electrons and protons:

At low pressures, atoms don’t break apart into individual electrons and protons. Instead, electrons are liberated from their atoms due to collisions with other particles. While the process involves ionization, it’s not a complete breakdown of the atom.
(d) The electrons in atoms can move freely at low pressures:

Although electrons do have more freedom of movement at low pressure, the key point is that this freedom is facilitated by the increased mean free path, leading to ionization. The ability of electrons to move freely alone is a consequence of the ionization process, not the direct cause of conductivity.

Hint

\(
\text { (a) } h \nu=\phi_0+E_{KE}=3.5+1.2=4.7 \mathrm{eV}
\)

Hint

(c) For doubly ionized helium \(\left(\mathrm{He}^{2+}\right)\), the charge \(Q_{H e}=2 e\) (where \(e\) is the elementary charge).
For hydrogen ion \(\left(\mathrm{H}^{+}\right)\), the charge \(Q_H=e\).

The work done (or energy gained) when a charge is accelerated through a potential difference \(V\) is given by:
\(
W=Q \cdot V
\)
For helium: \(W_{H e}=Q_{H e} \cdot V=2 e \cdot V\)
For hydrogen: \(W_H=Q_H \cdot V=e \cdot V\)
The work done on the particles is converted into kinetic energy. The kinetic energy \(K E\) is given by:
\(
K E=\frac{1}{2} m v^2
\)
For helium:
\(
\frac{1}{2} m_{H e} v_{H e}^2=2 e V
\)
For hydrogen:
\(
\frac{1}{2} m_H v_H^2=e V
\)
Express Velocities:
Rearranging the kinetic energy equations gives:
\(
\begin{aligned}
v_{H e}^2 & =\frac{4 e V}{m_{H e}} \\
v_H^2 & =\frac{2 e V}{m_H}
\end{aligned}
\)
Finding the Ratio of Velocities:
To find the ratio \(\frac{v_{H e}}{v_H}\) :
\(
\frac{v_{H e}^2}{v_H^2}=\frac{\frac{4 e V}{m_{H e}}}{\frac{2 e V}{m_H}}=\frac{4}{2} \cdot \frac{m_H}{m_{H e}}=2 \cdot \frac{m_H}{m_{H e}}
\)
Since the mass of helium \(m_{H e}\) is approximately 4 times the mass of hydrogen
\(
\begin{aligned}
& m_H \text { (i.e., } m_{H e}=4 m_H \text { ): } \\
& \frac{v_{H e}^2}{v_H^2}=2 \cdot \frac{m_H}{4 m_H}=\frac{2}{4}=\frac{1}{2}
\end{aligned}
\)
Taking the square root gives:
\(
\frac{v_{H e}}{v_H}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}
\)

Hint

\(
\begin{aligned}
& \text { (b) } W_0=\frac{h c}{\lambda_0} \text { or } W_0 \propto \frac{1}{\lambda_0} \\
& \Rightarrow \frac{W_1}{W_2}=\frac{\lambda_2}{\lambda_1}=\frac{600}{300}=2
\end{aligned}
\)

Hint

(a) Photoelectric current is directly proportional to the intensity of incident light.

Note: Kinetic energy of emitted photoelectrons varies with frequency of incident radiation and it does not depend upon the intensity of incident radiations. But, the number of photons emitted from the surface varies directly with intensity of incident light.

Hint

(a) Momentum of the photon \(p==\frac{h}{\lambda}=\frac{h \nu}{c}\)

\(\lambda=\frac{h}{m v}=\frac{h}{p}\)

Hint

(a) The work function has no effect on photoelectric current so long as \(h \nu>W_0\). The photoelectric current is proportional to the intensity of incident light. Since there is no change in the intensity of light, hence \(I_1=I_2\).

Hint

(d) Current \(=I_e+I_p \dots(i)\)
\(I_e\) and \(I_p\) are current due to electrons and positively charged ions.
\(
I=n e A V_d
\)
where,
\(
\begin{aligned}
n=5 \times 10^7 / \mathrm{cm}^3= & 5 \times 10^7 \times 10^6 / \mathrm{m}^3 \\
& =5 \times 10^{13} / \mathrm{m}^3
\end{aligned}
\)
\(
\begin{aligned}
& I_e=5 \times 10^{13} \times 1.6 \times 10^{-19} \times A \times 0.4 \\
& I_p=5 \times 10^{13} \times 1.6 \times 10^{-19} \times A \times v
\end{aligned}
\)
\(
\begin{aligned}
& I=I_e+I_p(\text { from equation (i)) } \\
& =5 \times 10^{13} \times 1.6 \times 10^{-19} \times A(v+0.4) \\
& \text { Given, } I / A=4 \times 10^{-6} A / \mathrm{m}^2 \\
& 4 \times 10^{-6} \times A=5 \times 10^{-6} \times 1.6 \times A(v+0.4) \\
& \frac{4}{8}=v+0.4 \Rightarrow 0.5=v+0.4 \Rightarrow v=0.1 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(b)

\(
\text { Here, } \frac{h c}{\lambda}=10^3 \text { ev and } h \nu=10^6 \mathrm{eV}
\)
Hence, \(\nu=\frac{10^3 \mathrm{c}}{\lambda}=\frac{10^3 \times 3 \times 10^8}{1.24 \times 10^{-9}}\) \(=2.4 \times 10^{20} \mathrm{~Hz}\)

Hint

(d)
\(
h \nu=W+\frac{1}{2} m v^2 \text { or } \frac{h c}{\lambda}=W+\frac{1}{2} m v^2
\)
\(
\begin{aligned}
&\begin{aligned}
& \text { Here } \lambda=3000 Å=3000 \times 10^{-10} \mathrm{~m} \\
& \text { and } W=1 \mathrm{eV}=1.6 \times 10^{-19} \text { joule } \\
& \begin{aligned}
\therefore \frac{\left(6.6 \times 10^{-34}\right)\left(3 \times 10^8\right)}{3000 \times 10^{-10}} \\
\quad=\left(1.6 \times 10^{-19}\right)+\frac{1}{2} \times\left(9.1 \times 10^{-31}\right) {v}^2
\end{aligned}
\end{aligned}\\
&\text { Solving we get } v \cong 10^6 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(c)

\(
m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{m_0}{\sqrt{\frac{c^2-(0.8 c)^2}{c^2}}}=\frac{5 m_0}{3}
\)

Hint

\(
\begin{aligned}
&\text { (a) No. of photons emitted per sec, }\\
&\begin{aligned}
& n=\frac{\text { Power }}{\text { Energy of photon }} \\
& =\frac{P}{h \nu}=\frac{10000}{6.6 \times 10^{-34} \times 880 \times 10^3}=1.72 \times 10^{31}
\end{aligned}
\end{aligned}
\)

Hint

(a) Momentum of the photon \(=\frac{h \mathrm{\nu}}{c}\)
\(
\begin{aligned}
\Rightarrow \frac{c}{\nu} & =\frac{h}{p}=\lambda \\
\nu & =\frac{c}{\lambda}=\frac{c p}{h}=3 \times 10^8 \times \frac{3.3 \times 10^{-29}}{6.6 \times 10^{-34}} \\
& =1.5 \times 10^{13} \mathrm{~Hz}
\end{aligned}
\)
where, \(\nu=\) frequency of radiation

Hint

(b) \(\quad\left(E_2-E_1\right)=h \nu=\frac{h c}{\lambda}\)
\(
\begin{aligned}
& \therefore \frac{h c}{\lambda_1}=\left(E_C-E_B\right), \frac{h c}{\lambda_2}=\left(E_B-E_A\right) \\
& \text { and } \frac{h c}{\lambda_3}=\left(E_C-E_A\right)
\end{aligned}
\)
Now,
\(
\begin{aligned}
& \left(E_C-E_A\right)=\left(E_C-E_B\right)+\left(E_B-E_A\right) \\
& \text { or, } \frac{h c}{\lambda_3}=\frac{h c}{\lambda_1}+\frac{h c}{\lambda_2} \text { or } \frac{1}{\lambda_3}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2} \\
& \therefore \frac{1}{\lambda_3}=\frac{\lambda_1+\lambda_2}{\lambda_1 \lambda_2} \text { or } \lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (b)  Kinetic energy of fastest electron }\\
&\begin{aligned}
& =E-W_0=6.2-4.2=2.0 \mathrm{eV} \\
& =2 \times 1.6 \times 10^{-19}=3.2 \times 10^{-19} \mathrm{~J}
\end{aligned}
\end{aligned}
\)

Hint

(a) \(\text { de Broglie wavelength, } \lambda=\frac{h}{p}=\frac{h}{m v}\)

Hint

\(
\text { (b) Energy of a photon } E=h \nu=\frac{h c}{\lambda}
\)

Hint

(b) Thermions are electrons, specifically electrons emitted from a heated material, a phenomenon known as thermionic emission.

Here’s a more detailed explanation:
Thermionic Emission: When a metal is heated, some of its electrons gain enough kinetic energy to overcome the surface barrier and escape, a process called thermionic emission.
Thermions: These emitted electrons are referred to as thermions.
Other options:
Protons: Protons are positively charged subatomic particles found in the nucleus of an atom.
Photons: Photons are particles of light, not related to thermionic emission.
Positrons: Positrons are the antimatter counterparts of electrons, having the same mass but a positive charge.

Hint

\(
\begin{aligned}
& \text { (a) }W_0=\frac{h c}{\lambda_0} \\
& =\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}}=4 \times 10^{-19} \mathrm{~J}
\end{aligned}
\)

Hint

(b) Photoelectric current is directly proportional to intensity of light.

Explanation: According to Einstein’s photoelectric equation, when light of sufficient frequency falls on a metal surface, it causes the emission of photoelectrons. The kinetic energy of the emitted electrons depends on the frequency of light, while the number of emitted electrons (and hence the photoelectric current) is directly proportional to the intensity of incident light.

This is because higher intensity means more photons striking the surface per second, leading to more electrons being emitted, which increases the current linearly as shown in the graph.

Hint

(c) 

\(r=0.052 n^2\)
For \(n=2\)
\(r=0.052 \times 4\)
\(=0.208 \mathrm{~nm}\)
\(M v r=\frac{n h}{2 \pi}\)
\(
\begin{aligned}
& \lambda=\frac{h}{M v}=\pi r \\
& =3.14 \times 0.208 \mathrm{~nm} \\
& =0.65317 \mathrm{~nm} \\
& \approx 0.67 \mathrm{~nm}
\end{aligned}
\)

Hint

(d) For photon, \(E=\frac{h c}{\lambda_{\mathrm{Ph}}} \Rightarrow \lambda_{\mathrm{Ph}}=\frac{h c}{E}\)
For electron, \(p=[latex] momentum and [latex]E=\frac{p^2}{2 m}=\left(\frac{h}{\lambda_e}\right)^2 \times \frac{1}{2 m}\)
\(
\begin{aligned}
& \Rightarrow \lambda_e=\frac{h}{\sqrt{2 m E}} \\
& \therefore \frac{\lambda_{\mathrm{Ph}}}{\lambda_e}=\frac{\frac{h c}{E}}{\frac{h}{\sqrt{2 m E}}}=c \sqrt{\frac{2 m}{E}}
\end{aligned}
\)