NEET Practice Q & A

Hint

(d) The maximum kinetic energy of the photo-electron is given by
\(
\Rightarrow \mathrm{K} \cdot \mathrm{E}_{\cdot \max }=\mathrm{h} \mathrm{f}-\phi
\)
where \(h\) is the Planck constant \(f\) is the frequency of the incident light or electromagnetic radiation. The term \(\phi\) is the work function.
Hence, \(\mathrm{KE}_{\text {max }}\) varies linearly with frequency and there are no terms representing the intensity of the incident light. 

Hint

\(
\begin{aligned}
&\begin{aligned}
& \text { (a) As, } \lambda=\frac{h}{p} \\
& \text { and } \lambda=\sqrt{\frac{h}{2 m E}} \\
& \Rightarrow \lambda \propto \frac{1}{p} \propto \frac{1}{E}
\end{aligned}\\
&\text { So, when wavelength is decreased, both } p \text { and } E \text { increase. }
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) As, } \lambda=\frac{h}{\sqrt{2 m K}} \\
& \Rightarrow \quad \lambda \propto \frac{1}{\sqrt{m}}
\end{aligned}
\)
Since, mass of \(\alpha\)-particle is largest as compare to all the particles given. So, \(\lambda\) is shortest for \(\alpha\)-particle.

Hint

\(
\begin{array}{ll}
\text { (a) As, } \lambda=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 q V m}} \\
\text { So, }  \lambda \propto \frac{1}{\sqrt{m}} \\
\therefore \frac{\lambda_e}{\lambda_p}=\left(\frac{m_p}{m_e}\right)^{1 / 2}
\end{array}
\)

Hint

(c) \(\begin{aligned} E(\text { in } \mathrm{eV})= & \frac{12375}{\lambda(\text { in } Å)} \\ & =\frac{12375}{7000}=1.76 \mathrm{eV}\end{aligned}\)

Hint

\(
\begin{array}{ll}
\text { (a) As, } p & =\frac{h v}{c} \\
\therefore \quad v & =\frac{p c}{h}=\frac{\left(6.6 \times 10^{-29}\right)\left(3 \times 10^8\right)}{\left(6.6 \times 10^{-34}\right)} \\
& =3 \times 10^{13} \mathrm{~Hz}
\end{array}
\)

Hint

(b) Given, \(E=100 \mathrm{eV}=\left(100 \times 1.6 \times 10^{-19}\right) \mathrm{J}\) and
\(
\begin{aligned}
h & =6.62 \times 10^{-34} \mathrm{Js} \\
E & =h \nu \\
\nu & =\frac{E}{h}=\frac{100 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}} \\
& =0.24 \times 10^{17}=2.417 \times 10^{16} \mathrm{~Hz}
\end{aligned}
\)

Hint

(c) According to Einstein’s photoelectric equation,
\(
\mathrm{KE}_{\max }=h \nu-\phi_0
\)
Comparing with the equation of straight line
\(
y=m x+c
\)
We get, slope of graph between KE and \(\nu=h\) (i.e. same for all metals)

Hint

\(
\begin{aligned}
&\text { (b) Work function, } W_0=h \nu_0\\
&\begin{array}{ll}
\text { For sodium, } & W_{01}=h \nu_{01}=\frac{h c}{\lambda_0} \\
\Rightarrow & \frac{2.3 \mathrm{eV}}{h}=v_{01}=\frac{c}{\lambda_{01}}
\end{array}
\end{aligned}
\)
\(
\begin{aligned}
&\text { For copper, } \quad W_{02}=h \nu_{02}\\
&\begin{array}{ll}
\Rightarrow & \frac{4.5 \mathrm{eV}}{h}={\nu}_{02}=\frac{c}{\lambda_{02}} \\
\Rightarrow & \frac{\lambda_{01}}{\lambda_{02}}=\frac{c h}{2.3} \times \frac{4.5}{c h} \simeq \frac{2}{1} \Rightarrow \lambda_{01}: \lambda_{02}=2: 1
\end{array}
\end{aligned}
\)

Hint

(a) According to Einstein’s photoelectric equation:
\(
K E_{\max }=h \nu-\phi
\)
The work function is also related to the threshold frequency \(\left(v_0\right)\) by the equation:
\(
\phi=h \nu_0
\)
Given the problem:
The incident light frequency is \(\boldsymbol{4} \boldsymbol{\nu}_0\).
The threshold frequency is \(\boldsymbol{\nu}_0\).
Substituting these values into the photoelectric equation:
\(
\begin{aligned}
& K E_{\max }=h\left(4 \nu_0\right)-h \nu_0 \\
& K E_{\max }=3 h \nu_0
\end{aligned}
\)

Hint

(a) As, \(\lambda=\frac{h}{p}=\frac{h}{m v}\)
So, \(\quad \frac{h}{m_1 v_1}=\frac{h}{m_2 v_2}\)
\(
\frac{v_1}{v_2}=\frac{m_2}{m_1}=\frac{4}{1} \quad\left(\because m_\alpha=4 m_p\right)
\)

Hint

(d) \(\begin{aligned} \text { As, } \lambda=\frac{h}{\sqrt{2 m E}} & =\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 80 \times 1.6 \times 10^{-19}}} \\ & =1.4 Å\end{aligned}\)

Hint

(c) (c) Momentum, \(p=\frac{h}{\lambda}\) or \(E=h \nu=\frac{h c}{\lambda}=p c\)
So, if the energy of photon is increased by a factor of 4 ,then its momentum will also increase by a factor of 4.

Hint

(a) Work function, \(W_0=h \nu_0\)
where, \(\nu_0=\) threshold frequency.
So,
\(
W \propto \nu_0
\)
Hence, \(\mathrm{Pt}>\mathrm{Al}>\mathrm{K}\)

Hint

\(
\begin{aligned}
&\text { (a) de-Broglie wavelength, }\\
&\begin{aligned}
\lambda & =\frac{h}{\sqrt{2 m k T}}=\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} T}} \\
& =\frac{6.62 \times 10^{-34}}{2.15 \times 10^{-25} \sqrt{T}} \mathrm{~m} \\
& =\frac{3.079}{\sqrt{T}} \times 10^{-9} \mathrm{~m}=\frac{30.79}{\sqrt{T}} \AA \approx \frac{30.8}{\sqrt{T}} Å
\end{aligned}
\end{aligned}
\)

Hint

(b) \(E=\frac{12375}{\lambda(Å)}=\frac{12375}{3500} \mathrm{eV}=3.53 \mathrm{eV}\)
Since, \(E>W_B\) but \(E<W_A\), therefore photoelectrons will be emitted by metal \(B\).

Hint

(d) Mass of an electron, \(m=9.11 \times 10^{-31} \mathrm{~kg}\)
\(
T=27+273=300 \mathrm{~K}
\)
de-Broglie wavelength of electron, \(\lambda=\frac{h}{\sqrt{3 m k T}}\)
\(
\begin{aligned}
& =\frac{6.6 \times 10^{-34}}{\sqrt{3 \times 9.11 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}} \mathrm{~m} \\
& =6.2 \times 10^{-9} \mathrm{~m}
\end{aligned}
\)

Hint

de-Broglie’s wavelength,
\(
\lambda=\frac{h}{m v}
\)
For electron,
\(
\lambda_e=\frac{h}{m_e v}
\)
For photon,
\(
\begin{aligned}
& \lambda_p=\frac{h}{m_p v} \\
& \frac{\lambda_e}{\lambda_p}=\frac{h / m_c v}{h / m_p v} \\
& \frac{\lambda_e}{\lambda_p}=\frac{m_p}{m_e} \\
& \frac{\lambda_e}{\lambda_p}=\frac{1.67 \times 10^{-27}}{9.1 \times 10^{-31}} \\
& =0.18 \times 10^4=1836
\end{aligned}
\)

Hint

(a) Wavelength of photon will be greater than that of electron because mass of photon is less than that of electron
\(
\Rightarrow \quad \lambda_{\mathrm{ph}}>\lambda_{\mathrm{el}} \quad\left(\because \lambda \propto \frac{1}{\sqrt{m}}\right)
\)

Hint

\(
\begin{aligned}
&\text { (c) Number of photons, }\\
&\begin{aligned}
n & =\frac{P}{E}=\frac{P}{h v}=\frac{P \lambda}{h c} \\
& =\frac{1000 \times 1986}{6.6 \times 10^{-34} \times 3 \times 10^8} \approx 10^{30}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\begin{aligned}
& \text { (b) } E=\frac{12375}{\lambda(\text { in } Å)}=\frac{12375}{4360}=2.84 \mathrm{eV} \\
& \therefore \quad K_{\max }=E-W=1.60 \mathrm{eV}
\end{aligned}\\
&\text { Thus, the retarding potential is } 1.60 \mathrm{~V} \text {. }
\end{aligned}
\)

Hint

(c)
\(
\begin{aligned}
& E_1=\frac{12375}{3500} \mathrm{eV}=3.54 \mathrm{eV} \\
& E_2=\frac{12375}{5400} \mathrm{eV}=2.29 \mathrm{eV}
\end{aligned}
\)
Now, \(\quad \frac{2}{1}=\frac{3.54-W}{2.29-W}\)
Solving this equation, we get \(W=1.05 \mathrm{eV}\).

Hint

(c) \(\frac{E_1}{E_2}=\frac{1}{2} \text { (given) }\)
We know that, \(\quad \lambda=\frac{h}{\sqrt{2 m K}} \Rightarrow \lambda \propto \frac{1}{\sqrt{K}}\)
\(
\begin{array}{ll}
\Rightarrow & \frac{\lambda_1}{\lambda_2}=\frac{\sqrt{K_2}}{\sqrt{K_1}} \Rightarrow \frac{\lambda_1^2}{\lambda_2^2}=\frac{K_2}{K_1} \\
\Rightarrow & \frac{\lambda_1^2}{\lambda_2^2}=\frac{2}{1} \Rightarrow \lambda_1: \lambda_2=\sqrt{2}: 1
\end{array}
\)

Hint

\(
\begin{aligned}
&\text { (b) Work function, } \phi_0=\frac{h c}{\lambda_0}\\
&\Rightarrow \quad \phi_0 \propto \frac{1}{\lambda_0} \Rightarrow \frac{\left(\phi_0\right)_1}{\left(\phi_0\right)_2}=\frac{\left(\lambda_0\right)_2}{\left(\lambda_0\right)_1}=\frac{600}{300}=2: 1
\end{aligned}
\)

Hint

(a) The change in wavelength is given by
\(
\Delta \lambda=\lambda^{\prime}-\lambda=\lambda_c(1-\cos \phi)
\)
where, \(\lambda^{\prime}=\) wavelength of the X-rays scattered and \(\lambda_c=\frac{h}{m_0 c}=\) compton wavelength \(=2.426 \times 10^{-12} \mathrm{~m}\)
\(
\Rightarrow \quad \lambda^{\prime}=\lambda+\lambda_c\left(1-\cos 30^{\circ}\right)=10 \mathrm{pm}+0.134 \lambda_c=10.32 \mathrm{pm}
\)

Hint

(c) By law of conservation of momentum,
\(
\begin{aligned}
0 & =m_1 v_1+m_2 v_2 \\
m_1 v_1 & =-m_2 v_2
\end{aligned}
\)
Negative sign indicates that both the particles are moving in opposite directions. Now, de-Broglie wavelengths,
\(
\begin{array}{ll}
& \lambda_1=\frac{h}{m_1 v_1} \text { and } \lambda_2=\frac{h}{m_2 v_2} \\
\therefore & \frac{\lambda_1}{\lambda_2}=\frac{m_2 v_2}{m_1 v_1}=1
\end{array}
\)

Hint

\(
\begin{aligned}
&\text { (a) Mass of electron in motion is }\\
&\begin{aligned}
m & =\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{m_0}{\sqrt{1-\frac{(0.5 c)^2}{c^2}}}=\frac{m_0}{\sqrt{0.75}} \\
\therefore \quad \lambda & =\frac{h}{m v}=\frac{6.6 \times 10^{-34} \times \sqrt{0.75}}{m_0 v} \\
& =\frac{6.6 \times 10^{-34} \times \sqrt{0.75}}{9.1 \times 10^{-31} \times 1.5 \times 10^8}=4.2 \times 10^{-12} \mathrm{~m}
\end{aligned}
\end{aligned}
\)

Hint

\(
\text { (b) } \quad K_1=h f-W \quad\left(\because W_1=W, f_1=f\right)
\)
\(
\text { and } K_2=2 h f-2 W \quad\left(\because W_2=2 W, f_2=2 f\right)
\)
\(
\therefore \quad \frac{K_1}{K_2}=\frac{1}{2}
\)

Hint

\(
\begin{array}{ll}
\text { (d) As, } \lambda & =\frac{h}{\sqrt{2 q V m}} \text { or } \lambda \propto \frac{1}{\sqrt{q V m}} \\
\therefore & \frac{\lambda_\alpha}{\lambda_p}=\frac{\sqrt{q_p V_p m_p}}{\sqrt{q_\alpha V_\alpha m_\alpha}}=\sqrt{\frac{1 \times 100 \times 1}{2 \times 800 \times 4}}=\frac{1}{8} \\
\therefore & \lambda_\alpha=\frac{\lambda_p}{8}=\frac{\lambda_0}{8}
\end{array}
\)

Hint

(b) de-Broglie wavelength, \(\lambda=\frac{h}{\sqrt{2 m E_e}}=\frac{h c}{E_p}\)
or \(2 m E_e=\frac{E_p^2}{c^2} \dots(i)\)
But \(E_e=\frac{1}{2} m v^2\) or \(m=\frac{2 E_e}{v^2}\)
From Eq. (i), we get
\(
2\left[\frac{2 E_e}{v^2}\right] E_e=\frac{E_p^2}{c^2} \text { or } \frac{4 E_e^2}{v^2}=\frac{E_p^2}{c^2}
\)
\(
\frac{E_e^2}{E_p^2}=\frac{v^2}{4 c^2} \quad \text { or } \quad \frac{E_e}{E_p}=\frac{v}{2 c}
\)

Hint

\(
\begin{aligned}
&\text { (c) From the given graph, we get }\\
&\begin{aligned}
V_0 & =-3.2 \mathrm{~V} \\
\text { As, } \quad e V_0 & =h v-\phi_0 \\
\Rightarrow \quad v & =\frac{e V_0+\phi_0}{h} \\
& =\frac{-3.2 \times 1.6 \times 10^{-19}+4 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} \\
& =1.939 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}
\end{aligned}
\)

Hint

(b) As,
\(
\begin{aligned}
E & =\frac{h c}{\lambda}-\phi_0 \dots(i)\\
4 E & =\frac{h c}{\lambda / 3}-\phi_0 \dots(ii)
\end{aligned}
\)
Solving these equations, we get \(\phi_0=\frac{h c}{3 \lambda}\)

Hint

\(
\begin{aligned}
\text { (a) } & e V_s & =h \nu-\phi \\
\therefore & V_s & =\left(\frac{h}{e}\right) \nu-\frac{\phi}{e}
\end{aligned}
\)
Comparing with equation of straight line \(y=m x+c\)
Slope \((m)=\tan \theta=\frac{h}{e}\)

Hint

\(
\begin{aligned}
&\text { (c) Maximum kinetic energy, }\\
&\begin{aligned}
K_{\max } & =\frac{h c}{\lambda}-\phi_0=\frac{6.4 \times 10^{-34} \times 3 \times 10^8}{6400 \times 10^{-10}}-1.6 \times 10^{-19} \\
& =1.4 \times 10^{-19} \mathrm{~J}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Work function, } \phi_0=\frac{h c}{\lambda_0}\\
&\begin{aligned}
\therefore & \frac{\left(\phi_0\right)_{\mathrm{T}}}{\left(\phi_0\right)_{\mathrm{Na}}} =\frac{\lambda_{\mathrm{Na}}}{\lambda_{\mathrm{T}}} \\
\text { or } & \lambda_{\mathrm{T}} =\frac{\lambda_{\mathrm{Na}} \times\left(\phi_0\right)_{\mathrm{Na}}}{\left(\phi_0\right)_{\mathrm{T}}}=\frac{5600 \times 2.3}{4.5}=2862 Å
\end{aligned}
\end{aligned}
\)

Hint

(d) According to Einstein’s photoelectric equation
\(
\mathrm{E}=\mathrm{W}_0+\mathrm{K}_{\max }
\)
\(
V_0=\frac{h c}{e}\left[\frac{1}{\lambda}-\frac{1}{\lambda_0}\right]
\)
So, \(V_0\) depends upon frequency and energy of incident radiation and nature of material .
Decreasing the wavelength of incident radiation, increases the stopping potential.
\(
\begin{aligned}
& \Rightarrow \mathrm{V}_2>\mathrm{V}_1 \\
& \Rightarrow \mathrm{~V}_2>2 \mathrm{~V}
\end{aligned}
\)

Hint

\(
\begin{aligned}
\text { (b) As, } \lambda & =\frac{h}{\sqrt{2 m E}} \\
\Rightarrow \lambda & \propto \frac{1}{\sqrt{m}} \\
\text { So, } \frac{\lambda_1}{\lambda_2} & =\sqrt{\frac{m_2}{m_1}} \\
\text { So, } \frac{\lambda}{\lambda_2} & =\sqrt{\frac{M}{m}}
\end{aligned}
\)
\(
\therefore \quad \lambda_2=\lambda \sqrt{\left(\frac{m}{M}\right)}
\)

Hint

(b) Intensity of incident light \(\propto\) No. of electrons emitted per sec
\(
\propto \frac{1}{(\text { distance })^2}
\)
\(\therefore\) If distance becomes \(d / 2\), then number of electrons becomes 4 times.

Hint

(b) Stopping potential, \(V_0=\frac{h}{e}\left(\nu-\nu_0\right)\)
\(
\begin{aligned}
\left(\nu-\nu_0\right) & =\frac{V_0 \times e}{h}=\frac{4.144 \times 1.6 \times 10^{-19}}{663 \times 10^{-34}} \\
\nu-1 \times 10^{15} & =10^{15} \\
\text { Frequency } & =10^{15}+1 \times 10^{15}=2 \times 10^{15} \mathrm{~Hz}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) Given, } v_{\max }=4 \times 10^8 \mathrm{~cm} / \mathrm{s}=4 \times 10^6 \mathrm{~m} / \mathrm{s} \\
& \therefore \quad K_{\max }=\frac{1}{2} m v_{\max }^2=\frac{1}{2} \times 9 \times 10^{-31} \times\left(4 \times 10^6\right)^2 \\
& =7.2 \times 10^{-18} \mathrm{~J}=45 \mathrm{eV}
\end{aligned}
\)
Hence, stopping potential, \(V_0=\frac{K_{\max }}{e}=\frac{45 \mathrm{eV}}{e}=45 \mathrm{~V}\)

Hint

(c) Wavelength \(\propto \frac{1}{\text { Energy }}\). Hence, for ejection of photoelectrons, wavelength should be smaller than threshold wavelength. As, wavelength of ultraviolet rays ( \(\sim 400 \mathrm{~nm}\) to 1 nm ) is less than that for infrared rays ( \(\sim 1 \mathrm{~nm}\) to 700 nm ). So, emission would be possible only ultraviolet lamp.

Hint

(b) When the distance is increased, frequency of incident light and hence the stopping potential does not change. But the intensity changes and hence the saturation current decreases nine times. \(\quad\left[\because\right.\)
Intensity of incident radiation \(\left.\propto \frac{1}{(\text { distance })^2}\right]\)
Cut-off voltage is independent of intensity and hence, remains the same.
Since distance becomes 3 times, intensity \((I)\) becomes \(\frac{I}{9}\).

Hint

(a) According to Einstein’s equation,
\(
h \nu=\phi_0+K_{\max }
\)
where, \(\phi_0\) is the work function.
\(
\Rightarrow \quad h \nu=\phi_0+e V_0 \Rightarrow V_0=\frac{h v}{e}-\frac{\phi_0}{e}
\)
Comparing with equation, \(y=m x+c\)
\(
\phi_0=- \text { (intercept) } \times e=O B \times e \text { in } \mathrm{eV}
\)

Hint

(d) If one electron enters a uniform electric field and another enters a uniform magnetic field, both moving at the same speed, their de Broglie wavelengths ( \(\lambda 1\) and \(\lambda 2\), respectively) will be related as follows: \(\lambda 1\) can be greater than, less than, or equal to \(\lambda 2\). This is because the electric field can change the electron’s speed, while the magnetic field only changes the direction of the electron’s motion, not its speed.

Speed of electron which enters into electric field may increase or decrease depending on direction of electric field while for second electron, since magnetic force is perpendicular to direction of velocity, it remains constant.

So, from \(\lambda=\frac{h}{m v}\) \(\lambda_1>\lambda_2\) or \(\lambda_1<\lambda_2\)

Hint

\(
\begin{aligned}
&\text { (b) Maximum kinetic energy, }\\
&K_{\max }(\operatorname{in} \mathrm{eV})=\frac{h c}{e}\left[\frac{1}{\lambda}-\frac{1}{\lambda_0}\right]=12375\left[\frac{1}{1000}-\frac{1}{2000}\right]=6.2 \mathrm{eV}
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
\text { (c) de-Broglie wavelength, } \lambda=\frac{h}{p} \\
\Rightarrow \lambda-\frac{0.5 \lambda}{100}=\frac{h}{p+\Delta p} \\
\Rightarrow \frac{199 \lambda}{200}=\frac{h}{p+\Delta p}=\frac{199 h}{200 p} \\
\Rightarrow p+\Delta p=\frac{200}{199} p \\
\Rightarrow p=199 \Delta p
\end{array}
\)

Hint

\(
\begin{array}{ll}
\text { (a) Here, } & \lambda_1=\frac{12375}{E_1(\mathrm{eV})} Å=1000 Å \\
\therefore E_1=12.375 \mathrm{eV} \\
\text { Similarly, } E_2(\mathrm{eV}) & =\frac{12375}{\lambda_2(\AA)} \mathrm{eV}=\frac{12375}{2000}=6.1875 \mathrm{eV}
\end{array}
\)
Now, \(\quad E_1-W_0=e V_1\)
and \(\quad E_2-W_0=e V_2\)
Hence, \(12.375-W_0=7.7 \mathrm{eV}\)
and \(\quad 6.1875-W_0=e V_2\)
Solving, we get \(V_2=1.5 \mathrm{~V}\)

Hint

(b) According to the question,
\(
\begin{aligned}
& e V=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) \dots(i) \\
& \frac{e V}{3}=h c\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right) \dots(ii)
\end{aligned}
\)
Dividing Eq. (i) by Eq. (ii), we get
\(
\begin{aligned}
3 & =\frac{\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)}{\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right)} \\
3\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right) & =\frac{1}{\lambda}-\frac{1}{\lambda_0} \\
\frac{3}{2 \lambda}-\frac{1}{\lambda} & =\frac{3}{\lambda_0}-\frac{1}{\lambda_0} \\
\frac{1}{2 \lambda} & =\frac{2}{\lambda_0}
\end{aligned}
\)
Threshold wavelength for metallic surface, \(\lambda_0=4 \lambda\)

Hint

(a) Energy radiated as visible light \(=\frac{5}{100} \times 100=5 \mathrm{~J} / \mathrm{s}\)
Let \(n\) be the number of photons emitted per second, then
\(
\begin{aligned}
n h v & =E=5 \\
\therefore \quad n & =\frac{5 \lambda}{h c}=\frac{5 \times 5.6 \times 10^{-7}}{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^8\right)} \\
& \approx 1.4 \times 10^{19}
\end{aligned}
\)

Hint

(c) Intensity, \(I=10^{-10} \mathrm{Wbm}^{-2}=10^{-10} \mathrm{Js}^{-1} \mathrm{~m}^{-2}\).
Let the number of photons required be \(n\), then
\(
\begin{aligned}
I & =\frac{n h v}{A}=\frac{n h v}{10^{-4}}=10^{-10} \\
n & =\frac{10^{-10} \times 10^{-4}}{h v}=10^{-14} \frac{\lambda}{h c} \\
& =\frac{10^{-14} \times 660 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8}=3.3 \times 10^4
\end{aligned}
\)

Hint

(a) As \(\frac{m v^2}{r}=q v B \Rightarrow \frac{m v}{q B}=r\)
Also, \(m v=p \Rightarrow p=\sqrt{2 m K}\)
\(
\Rightarrow \quad r=\frac{\sqrt{2 K m}}{B q}=\frac{\sqrt{2\left(E-W_0\right) m}}{e B}
\)

Hint

(b) As, \(E-W_0=\frac{1}{2} m v^2\)
First case, \(\quad 2 W_0-W_0=\frac{1}{2} m v_1^2 \dots(i)\)
Second case, \(\quad 5 W_0-W_0=\frac{1}{2} m v_2^2 \dots(ii)\)
Therefore, from Eqs. (i) and (ii), we get
\(
\frac{2 W_0-W_0}{5 W_0-W_0}=\frac{v_1^2}{v_2^2} \text { or } \frac{v_1}{v_2}=\frac{1}{2}
\)

Hint

(b) Energy corresponding to \(2000 Å\),
\(
E=\frac{12375}{\lambda(\sim \operatorname{in} Å)}=\frac{12375}{2000} \mathrm{eV}=6.2 \mathrm{eV}
\)
Maximum kinetic energy,
\(
\begin{aligned}
K & =h v-h v_0 \\
K & =E-W \\
& =(6.2-5.01) \mathrm{eV}=1.19 \mathrm{eV}
\end{aligned}
\)
Now,
\(
K=\frac{1}{2} m v_{\max }^2
\)
\(
\Rightarrow \frac{1}{2} \times 9.1 \times 10^{-31} \times v_{\max }^2=1.19 \times 1.6 \times 10^{-19}
\)
\(
\begin{aligned}
& v_{\max }^2=\frac{1.19 \times 1.6 \times 10^{-19} \times 2}{9.1 \times 10^{-31}} \\
& v_{\max }^2=0.418 \times 10^{12}=41.8 \times 10^{10} \\
& v_{\max }=6.46 \times 10^5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(b) From equation,
\(
\begin{aligned}
e V_0 & =h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) \\
4.8 & =\frac{h c}{e}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) \dots(i) \\
1.6 & =\frac{h c}{e}\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right) \dots(ii)
\end{aligned}
\)
From Eqs. (i) and (ii), we get
\(
\frac{\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)}{\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right)}=\frac{4.8}{1.6} \Rightarrow \lambda_0=4 \lambda
\)

Hint

\(
\begin{aligned}
& \text { (b) As, } \lambda=\frac{h}{\sqrt{2 m K}} \text { or } \lambda^2=\frac{h^2}{2 m K} \\
& \begin{aligned}
\therefore \quad K & =\frac{h^2}{2 m \lambda^2}=\frac{\left(6.6 \times 10^{-34}\right)^2}{2 \times 9.1 \times 10^{-31} \times\left(10 \times 10^{-12}\right)^2} \mathrm{~J} \\
& =15.1 \mathrm{keV} \simeq 15 \mathrm{eV}
\end{aligned}
\end{aligned}
\)

Hint

(a) Consider a spherical chamber of radius 20 Bulb is placed at point \(S\) as shown in the figure.

Intensity of light at surface,
\(
I=\frac{0.6 P}{4 \pi r^2}
\)
Force exerted on the surface, \(F=\frac{P}{c}\), where \(P\) is power of bulb.
\(
\therefore \quad F=\frac{1100}{3 \times 10^8}=3.6 \times 10^{-6} \mathrm{~N}
\)
\(
\begin{aligned}
\text { Pressure exerted by light } & =\frac{I}{c}=\frac{0.6 P}{4 \pi r^2 c}=\frac{0.6 \times 1100}{4 \pi(0.2)^2 \times 3 \times 10^8} \\
& =4.4 \times 10^{-6} \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Force totally reflecting surface, }\\
&\begin{aligned}
F & =\frac{2 P \cos 60^{\circ}}{c} \\
& =\frac{2 \times \frac{n h c}{\lambda} \cos 60^{\circ}}{c}=\frac{10^{19} \times 6.6 \times 10^{-34}}{663 \times 10^{-9}}=10^{-8} \mathrm{~N}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) As, }\\
&\begin{aligned}
\frac{1}{2} m v_{\max }^2 & =\frac{h c}{\lambda}-\phi_0 \\
v_{\max } & =\sqrt{\frac{2}{m}\left(\frac{h c}{\lambda}-\phi_0\right)}=v \\
v_{\max }^{\prime} & =\sqrt{\frac{2}{m}\left(\frac{h c}{\lambda / 4}-\phi_0\right)} \\
& =2 \sqrt{\frac{2}{m}\left(\frac{h c}{\lambda}-\phi_0\right)+\frac{6}{m} \phi_0} \\
& >2 v
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Energy of photoelectron, }\\
&\Rightarrow \quad \begin{array}{ll}
E & =\frac{h c}{\lambda}-\phi_0 \\
\Rightarrow & \frac{h c}{\lambda}
\end{array}=E+\phi_0 \dots(i)
\end{aligned}
\)
where \(\phi_0\) is the work function for the metal surface (constant). When \(E^{\prime}=2 E\), then
\(
\begin{aligned}
& 2 E & =\frac{h c}{\lambda^{\prime}}-\phi_0 \\
\Rightarrow & \frac{h c}{\lambda^{\prime}} & =2 E+\phi_0 \dots(ii)
\end{aligned}
\)
Dividing Eq. (i) by Eq. (ii), we get
\(
\frac{\lambda^{\prime}}{\lambda}=\frac{E+\phi_0}{2 E+\phi_0}
\)
\(
\begin{aligned}
\frac{\lambda^{\prime}}{\lambda} & =\frac{\left(E+\phi_0\right)}{2\left(E+\frac{\phi_0}{2}\right)} \\
\frac{\lambda^{\prime}}{\lambda} & >\frac{1}{2} \text { or } \lambda^{\prime}>\frac{\lambda}{2} \\
\lambda & >\lambda^{\prime}>\frac{\lambda}{2}
\end{aligned}
\)

Hint

(c) Let \(W\) be the work function of metal, then
\(
\begin{aligned}
e V_0 & =\frac{h c}{330 \times 10^{-9}}-W \dots(i) \\
e\left(2 V_0\right) & =\frac{h c}{220 \times 10^{-9}}-W \dots(ii)
\end{aligned}
\)
Solving these two equations, we get
\(
\begin{aligned}
V_0 & =\frac{10^9 \times h \times c}{110 \times e \times 6} \\
& =\frac{10^9 \times 6.6 \times 10^{-34} \times 3 \times 10^8}{110 \times 1.6 \times 10^{-19} \times 6}=\frac{15}{8} \mathrm{~V}
\end{aligned}
\)

Hint

(c) As, \(e V=\frac{h c}{\lambda}-W\)
\(
\therefore \quad V=\left(\frac{h}{e}\right) \cdot \frac{c}{\lambda}-\frac{W}{e}
\)
So,
\(
\begin{aligned}
V_1 & =\left(\frac{h}{e}\right) \frac{c}{\lambda_1}-\frac{W}{e} \dots(i) \\
V_2 & =\left(\frac{h}{e}\right) \frac{c}{\lambda_2}-\frac{W}{e} \dots(ii)
\end{aligned}
\)
Solving these two equations, we get
\(
\begin{aligned}
\frac{h}{e} & =\frac{\lambda_1 \lambda_2\left(V_1-V_2\right)}{c\left(\lambda_2-\lambda_1\right)} \\
& =\frac{\left(0.6 \times 0.4 \times 10^{-12}\right)(1.0)}{\left(3 \times 10^8\right)\left(0.2 \times 10^{-6}\right)} \\
& =4 \times 10^{-15} \mathrm{Vs}
\end{aligned}
\)

Hint

(d) We know, \(K_{\max }=E-W\)
\(
\begin{array}{ll}
\therefore & T_A=4.25-W_A \dots(i) \\
& T_B=\left(T_A-1.50\right)=4.70-W_B \dots(ii)
\end{array}
\)
From these two equations, we have
\(
W_B-W_A=1.95 \mathrm{eV} \dots(iii)
\)
de-Broglie wavelength is given by
\(
\begin{aligned}
\lambda & =\frac{h}{\sqrt{2 K m}} \text { or } \lambda \propto \frac{1}{\sqrt{K}} \\
\frac{\lambda_B}{\lambda_A} & =\sqrt{\frac{K_A}{K_B}} \Rightarrow 2=\sqrt{\frac{T_A}{T_A-1.5}} \\
T_A & =2 \mathrm{eV}, W_A=2.25 \mathrm{eV}, W_B=4.20 \mathrm{eV} \\
T_B & =0.5 \mathrm{eV}
\end{aligned}
\)

Hint

(c) As we know that, wavelength of a charged particle,
\(
\lambda=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m}} \cdot \frac{1}{\sqrt{E}}
\)
\(
\begin{aligned}
& \log \lambda=\log \left(\frac{h}{\sqrt{2 m}} \cdot \frac{1}{\sqrt{E}}\right) \\
& \log \lambda=\log \frac{h}{\sqrt{2 m}}+\log \frac{1}{E^{1 / 2}} \\
& \log \lambda=\log \left(\frac{h}{\sqrt{2 m}}\right)-\frac{1}{2} \log E \\
& \log \lambda=-\frac{1}{2} \log E+\log \frac{h}{\sqrt{2 m}}
\end{aligned}
\)
Comparing the above equation with the equation of straight line, i.e. \(\quad y=m x+c\).
We can conclude that, \(\log \lambda\) varies linearly with \(\log E\), with slope \(-1 / 2\).
It has been correctly shown in graph (c).

Hint

(d) The de-Broglie wavelength of an electron,
\(
\lambda=\frac{h}{m v} \Rightarrow v=\frac{h}{m \lambda} \dots(i)
\)
Energy of photon,
\(
E_p=\frac{h c}{\lambda} \text { (since, } \lambda \text { is same for electron and photon) } \dots(ii)
\)
Kinetic energy of photon to kinetic energy of electron
\(
=\frac{E_p}{E_e}=\frac{\frac{h c}{\lambda}}{\left(\frac{1}{2}\right) m v^2}=\frac{2 h c}{\lambda m v^2}
\)
Substituting the value of \(v\) from Eq. (i), we get
\(
\frac{E_p}{E_e}=\frac{2 h c}{\lambda m\left(\frac{h}{m \lambda}\right)^2}=\frac{2 \lambda m c}{h}
\)

Hint

(b) When charged particle is moving on a circular path in a uniform magnetic field \(B\), then
\(
\frac{m v^2}{r}=q v B \Rightarrow \frac{m v}{r}=q B \Rightarrow v=\frac{q B r}{m}
\)
Thus, kinetic energy \(=\frac{1}{2} m v^2=\frac{1}{2} m \times\left(\frac{q B r}{m}\right)^2=\frac{1}{2 m}(q B r)^2\)
Now for photoelectric effect, Einstein’s equation is
\(
h \nu=W+(\mathrm{KE})
\)
where,
\(
W=\text { work function }
\)
\(
\Rightarrow \quad h \nu-\mathrm{KE}=W \Rightarrow W=h v-\frac{(q B r)^2}{2 m}
\)
Here, charged particle is a photoelectron denoted by \(e\)
\(
\therefore \quad W=h \nu-\frac{(e B r)^2}{2 m_e}
\)

Hint

(b) Given that, \(2 \mathrm{eV} \leq \phi \leq 5 \mathrm{eV}\)
Wavelength corresponding to minimum and maximum values of work function are
\(
\lambda_{\max }=\frac{h c}{E}=\frac{4 \times 10^{-15} \times 3 \times 10^8}{2}
\)
\(
=6 \times 10^{-7} \mathrm{~m}=600 \mathrm{~nm}
\)
and
\(
\begin{aligned}
\lambda_{\min } & =\frac{h c}{E}=\frac{4 \times 10^{-15} \times 3 \times 10^8}{5} \\
& =2.4 \times 10^{-7} \mathrm{~m}=240 \mathrm{~nm}
\end{aligned}
\)
Clearly, wavelength range would be
\(
240 \mathrm{~nm} \leq \lambda \leq 600 \mathrm{~nm}
\)
Thus, wavelength 650 nm is not suitable for photoelectric effect.

Hint

(d) The de-Broglie wavelength is
\(
\lambda=\frac{h}{m v} \Rightarrow m=\frac{h}{\lambda v} \Rightarrow m \propto \frac{1}{\lambda}, \text { if } \frac{h}{v}=\text { constant }
\)
Again, \(\frac{h}{v}\) will be constant only when \(v\) is constant.
Thus, in present case the de-Broglie wavelength of heavier particle will be smaller, when the velocities of two particles will be same.

Hint

(a) The de-Broglie wavelength of electron associated with voltage is given by
\(
\lambda=\frac{1.227}{\sqrt{V}} \mathrm{~nm}=\frac{1.227}{\sqrt{64}} \mathrm{~nm}=0.153 \mathrm{~nm}
\)

Hint

(d) Only if the frequency of the incident radiation is above a certain threshold value.

Explanation:
The photoelectric effect describes the phenomenon where electrons are emitted from a metal surface when light shines on it. However, this emission only occurs if the frequency of the light is greater than a specific threshold frequency for the metal. This threshold frequency is determined by the work function of the metal, which is the minimum energy required to free an electron from the surface. If the light’s frequency is below this threshold, the electrons do not have enough energy to escape the metal.

Hint

(b) From Einstein’s photoelectric equation, \((\mathrm{KE})_{\max }=h \nu-\phi_0\)
For first condition, \(0.5=E-\phi_0 \dots(i)\)
For the second condition,
\(
\begin{aligned}
0.8 & =\left(E+\frac{E \times 20}{100}\right)-\phi_0 \\
\Rightarrow \quad 0.8 & =1.2 E-\phi_0 \dots(ii)
\end{aligned}
\)
From Eqs. (i) and (ii), we get
\(
\begin{array}{rlrl}
0.3 & =0.2 E \\
\Rightarrow & E =\frac{3}{2}=1.5 \mathrm{eV}
\end{array}
\)
From Eq. (i), \(\phi_0=E-0.5=1.5-0.5=1 \mathrm{eV}\)

Hint

\(
\begin{aligned}
&\text { (a) According to Einstein’s photoelectric equation, }\\
&\begin{aligned}
& K_A=\frac{h c}{\lambda_A}-\phi \Rightarrow K_A+\phi=\frac{h c}{\lambda_A} \\
& K_B=\frac{h c}{\lambda_B}-\phi=\frac{h c}{\left(\lambda_A / 2\right)}-\phi \\
& K_B=\frac{2 h c}{\lambda_A}-\phi=2\left(K_A+\phi\right)-\phi=2 K_A+\phi \Rightarrow K_A<\frac{K_B}{2}
\end{aligned}
\end{aligned}
\)

Hint

(b) Above the threshold frequency, the maximum kinetic energy of the emitted photoelectrons depends on the frequency of the incident light, but is independent of charge, pressure, angle of incidence, etc., of the incident light.

Note: \(\text { K.E.max }={h\nu}-\phi\)

Hint

(c) From photoelectric equation, \(E_K=E-W\)
where, \(E_K\) is kinetic energy of emitted photoelectrons, \(W\) is the work function and \(E\) is the energy supplied.
\(
E=h \nu=\frac{h c}{\lambda}
\)
where, \(h\) is Planck’s constant, \(c\) is the speed of light and \(\lambda\) is the wavelength.
\(
\begin{array}{ll}
\therefore & E=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}}=3.96 \times 10^{-19} \mathrm{~J} \\
\text { Also, } & 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J} \\
\therefore & E=\frac{3.96 \times 10^{-19}}{1.6 \times 10^{-19}}=2.48 \mathrm{eV} \\
\text { Hence, } & E_K=2.48-1.90=0.58 \mathrm{eV}
\end{array}
\)

Hint

(a) We should know that the maximum kinetic energy of emitted photoelectrons is given by
\(
K_{\max }=h \nu-\phi
\)
In the above formula,
\(K_{\text {max }}\) represents the maximum kinetic energy
h represents the Planck’s constant
and \(\nu\) is represents the frequency
Hence, it is dependent mainly on the energy (that is frequency) of the incident photon. On doubling the intensity, \(K_{\max }\) remains the same.

Explanation: Maximum kinetic energy of the photoelectrons, does not depend upon the intensity of the incident light. If the intensity of light be doubled, then the number of photoelectrons emitted per second is doubled. So, \(n\) is doubled but \(K_{\max }\) remains same.

Hint

(a) The photoelectric emission is an instantaneous process without any apparent time lag.
Below threshold frequency no electron can be emitted.
According to Einstein.s photoelectric equation,
\(
K_{\max }=h \nu-\phi_0
\)
where \(\nu\) is the frequency of incident radiation if it is more then the threshold frequency.

The maximum kinetic energy of the emitted photoelectrons is proportional to the frequency of incident radiation.

Hint

(d) If the wavelength of incident light falling on a photosensitive meterial decreases, then stopping potential increases. This is because
\(
V_0=\frac{h}{e}\left(\nu-\nu_0\right) \Rightarrow V_0=\frac{h}{e}\left(\frac{c}{\lambda}-\frac{c}{\lambda_0}\right) \Rightarrow V_0 \propto \frac{1}{\lambda}
\)

Hint

(c) de-Broglie wavelength, \(\lambda=\frac{h}{\sqrt{2 m K}}\)
The kinetic energy of the electron, \(K_{\text {electron }}=\frac{h^2}{2 m \cdot \lambda^2} \dots(i)\)
where, \(h=\) Planck’s constant, \(\lambda=\) wavelength
The photon energy, \(E_{\text {photon }}=\frac{c h}{\lambda} \dots(ii)\)
From Eqs. (ii) and (i), we get
\(
\begin{aligned}
\frac{E_{\text {photon }}}{K_{\text {electron }}} & =\frac{c h / \lambda}{h^2 / 2 m \cdot \lambda^2}=\frac{h c \cdot \lambda^2 2 m}{h^2 \cdot \lambda}=\frac{2 m \lambda c}{h} \\
& =\frac{2\left(m c^2\right)}{h c / \lambda}=\frac{2 \times\left(5 \times 10^5\right)}{50 \times 10^3}=20 \\
F_{\text {photon }}: K_{\text {electron }} & =20: 1
\end{aligned}
\)

Hint

(c) Given, \(v=1.8 \times 10^6 \mathrm{~ms}^{-1}\)
\(
\frac{e}{m}=1.8 \times 10^{11} \mathrm{Ckg}^{-1}
\)
We have, \(\quad e V_0=\frac{1}{2} m v^2 \Rightarrow V_0 \frac{e}{m}=\frac{v^2}{2}\)
\(
\Rightarrow \quad V_0 \times 1.8 \times 10^{11}=\frac{1.8 \times 1.8 \times\left(10^6\right)^2}{2} \Rightarrow V_0=9 \mathrm{~V}
\)

Hint

(a) de-Broglie wavelength is given by
So,
\(
\begin{aligned}
& \lambda=\frac{h}{p} \text { and } \frac{h}{p} \propto \frac{1}{\sqrt{K}} \\
& \lambda \propto \frac{1}{\sqrt{K}} \text { or } \lambda \sqrt{K}=\mathrm{constant}
\end{aligned}
\)
Hence, the graph between \(\lambda\) and \(K\) will be a curved line.

Hint

(d) A photocell converts light energy into electrical energy.

Explanation:
A photocell, also known as a photovoltaic cell, works by using the photoelectric effect. When light strikes the cell, it excites electrons in a semiconductor material, causing them to flow and generate an electric current. This process converts the light energy into electrical energy.

Hint

(d) Mmaximum energy of an electron will be \(E=\frac{h c}{\lambda}-\phi\)
Where \(\phi\) is the work function and ‘ \(h\) ‘ is the planck’s constant and \(\lambda\) is the wavelength of the light and ‘ \(c\) ‘ is the velocity of light and ‘ \(E\) ‘ is the maximum kinetic energy.
Work function will be constant for a metal and as the wavelength of light decreases the energy of radiation increases and the maximum kinetic energy of the photoelectron emitted increases.

Hint

(b) Photoelectric current depends on the intensity of the incident light and is independent of the frequency of the incident light.

Explanation:
The photoelectric current, which is the flow of electrons emitted from a material when light shines on it, is directly proportional to the number of photons that strike the material per unit time. This number of photons is determined by the intensity of the light beam. More intense light means more photons, which leads to more electrons being ejected and a higher photoelectric current.

Hint

\(
\begin{array}{ll}
\text { (c) Momentum, } p=m v=\frac{h}{\lambda} \text { or } v=\frac{h}{m \lambda} \\
\therefore v=\frac{6.62 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.2 \times 10^{-7}} \\
\Rightarrow v=\frac{6.62 \times 10^4}{9.1 \times 5.2} \Rightarrow v \cong 1400 \mathrm{~ms}^{-1}
\end{array}
\)

Hint

\(
\begin{aligned}
&\text { (a) }\\
&\begin{gathered}
K_e=\frac{1}{2} m v^2 \text { and } \lambda=\frac{h}{m v} \\
K_e=\frac{1}{2}\left(\frac{h}{\lambda v}\right) v^2=\frac{v h}{2 \lambda} \text { and } E_p=\frac{h c}{\lambda}
\end{gathered}
\end{aligned}
\)
\(
\therefore \quad \frac{K_e}{E_p}=\frac{v}{2 c}=\frac{1.5 \times 10^8}{2 \times 3 \times 10^8}=\frac{1}{4}
\)

Hint

(a) According to Einstein’s photoelectric equation,
\(
K_{\max }=h \nu-h \nu_0
\)
Initially, \(\frac{1}{2} m v_0^2=h\left(2 \nu_0\right)-h\left(\nu_0\right)=h \nu_0 \dots(i)\)
Then, when, \(\nu=5 \nu_0\)
\(
\begin{aligned}
& & \frac{1}{2} m v^2 & =5 h \nu_0-h \nu_0=4 h \nu_0 \\
\Rightarrow & & \frac{1}{2} m v^2 & =4\left(1 / 2 m v_0^2\right) [\because \text { from Eq. (i) }]\\
\Rightarrow & & v^2 & =4 v_0^2 \text { or } v=\sqrt{2} v_0
\end{aligned}
\)

Hint

(c) The process of emitting electrons from a hot surface is known as thermionic emission. When a metal is heated sufficiently, electrons gain enough thermal energy to overcome the material’s work function – the minimum energy required for an electron to escape from the surface. These emitted electrons are often referred to as thermions.
Why the other options are incorrect:
(a) Nuclei: Nuclei contain protons and neutrons and are much more tightly bound within the atom than electrons. It would require significantly higher energy to remove them from the surface.
(b) Neutrons: Similar to protons, neutrons are part of the nucleus and are not easily ejected by heat alone.
(d) Protons: Protons, also within the nucleus, are not released in this process.