Einstein's Photoelectric Equation and Energy Quantum of Radiation Questions in English

(A) The work function $W_0$ is given by the formula $W_0 = \frac{hc}{\lambda_0}$.
Using the approximation $hc \approx 12375 \, eV \cdot \mathring{A}$,we have:
$W_0 = \frac{12375}{6800} \, eV$.
$W_0 \approx 1.82 \, eV$.
Rounding to the nearest given option,the work function is $1.8 \, eV$.

(C) The energy of the incident photon is given by $E = h\nu$.
Given $h = 6.626 \times 10^{-34} \ J \cdot s$ and $\nu = 8 \times 10^{15} \ Hz$.
$E = 6.626 \times 10^{-34} \times 8 \times 10^{15} \approx 5.3 \times 10^{-18} \ J$.
To convert this energy into electron-volts $(eV)$,divide by $1.6 \times 10^{-19} \ J/eV$:
$E = \frac{5.3 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 33.125 \ eV$.
Using Einstein's photoelectric equation: $K_{\max} = E - W_0$.
Given work function $W_0 = 6.125 \ eV$.
$K_{\max} = 33.125 \ eV - 6.125 \ eV = 27 \ eV$.

(B) The threshold wavelength $\lambda_0$ is given by the formula $\lambda_0 = \frac{hc}{W_0}$.
Given the work function $W_0 = 2 \ eV$.
Using the relation $\lambda_0 (\text{in } \mathring{A}) \approx \frac{12400}{W_0 (\text{in } eV)}$.
$\lambda_0 = \frac{12400}{2} = 6200 \ \mathring{A}$.
Since $1 \ nm = 10 \ \mathring{A}$,we have $\lambda_0 = 620 \ nm$.
Therefore,the correct option is $B$.

(A) According to Einstein's photoelectric equation, the maximum kinetic energy $(K_{max})$ of emitted photo-electrons is given by $K_{max} = h\nu - \Phi$, where $h\nu$ is the energy of the incident photon and $\Phi$ is the work function of the metal.
Given: $h\nu = 6 eV$ and $\Phi = 4 eV$.
$K_{max} = 6 eV - 4 eV = 2 eV$.
The photo-electrons are emitted with a range of kinetic energies starting from $0$ up to the maximum kinetic energy $K_{max}$.
Therefore, the minimum kinetic energy of the emitted photo-electrons is $0 eV$.

(C) The minimum frequency required for photoelectric emission is called the threshold frequency $(\nu_0)$.
It is related to the work function $(W_0)$ by the equation: $W_0 = h\nu_0$.
Given: $W_0 = 1.65 \ eV = 1.65 \times 1.6 \times 10^{-19} \ J$ and Planck's constant $h = 6.6 \times 10^{-34} \ J \cdot s$.
Substituting these values into the equation:
$\nu_0 = \frac{W_0}{h} = \frac{1.65 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}}$
$\nu_0 = \frac{2.64 \times 10^{-19}}{6.6 \times 10^{-34}} = 0.4 \times 10^{15} \ Hz = 4 \times 10^{14} \ Hz$.
Therefore,the correct option is $C$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = E - W_0$,where $E$ is the energy of the incident photon and $W_0$ is the work function of the metal.
For the first photon with energy $E_1 = 1 \text{ eV}$:
$K_1 = E_1 - W_0 = 1 \text{ eV} - 0.5 \text{ eV} = 0.5 \text{ eV}$.
For the second photon with energy $E_2 = 2.5 \text{ eV}$:
$K_2 = E_2 - W_0 = 2.5 \text{ eV} - 0.5 \text{ eV} = 2.0 \text{ eV}$.
The ratio of the maximum kinetic energies is:
$\frac{K_1}{K_2} = \frac{0.5}{2.0} = \frac{1}{4}$.
Thus,the ratio is $1 : 4$.

(C) The work function $W_0$ is related to the threshold wavelength $\lambda_0$ by the formula $W_0 = \frac{hc}{\lambda_0}$.
From this,we can see that $W_0 \propto \frac{1}{\lambda_0}$,which implies $\lambda_0 \propto \frac{1}{W_0}$.
Given the work functions for sodium $(W_1 = 2.3 \ eV)$ and copper $(W_2 = 4.5 \ eV)$,the ratio of their threshold wavelengths is:
$\frac{\lambda_1}{\lambda_2} = \frac{W_2}{W_1} = \frac{4.5 \ eV}{2.3 \ eV} \approx \frac{4.6}{2.3} = 2$.
Thus,the ratio is $2:1$.

(D) The relationship between the maximum kinetic energy $(K_{\max})$ of photoelectrons and the stopping potential $(V_0)$ is given by the equation: $K_{\max} = eV_0$.
Given that the maximum kinetic energy $K_{\max} = 4.0 \ eV$.
Substituting this into the equation: $4.0 \ eV = eV_0$.
Therefore,the stopping potential $V_0 = 4.0 \ V$.

(C) The maximum kinetic energy of the ejected photoelectrons is related to the stopping potential $V_0$ by the equation: $K_{\max} = \frac{1}{2}mv_{\max}^2 = eV_0$.
Rearranging this equation to solve for the maximum velocity $v_{\max}$,we get: $v_{\max} = \sqrt{2(\frac{e}{m})V_0}$.
Given values: $V_0 = 9 \ V$ and $\frac{e}{m} = 1.8 \times 10^{11} \ C \ kg^{-1}$.
Substituting these values into the formula: $v_{\max} = \sqrt{2 \times (1.8 \times 10^{11}) \times 9}$.
$v_{\max} = \sqrt{32.4 \times 10^{11}} = \sqrt{3.24 \times 10^{12}} = 1.8 \times 10^6 \ m \ s^{-1}$.

(B) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + K_{\max}$.
For plate $A$: $\frac{hc}{\lambda_A} = W_0 + K_A$ ... $(i)$
For plate $B$: $\frac{hc}{\lambda_B} = W_0 + K_B$ ... $(ii)$
Given ${\lambda _A} = 2{\lambda _B}$,substitute this into $(i)$:
$\frac{hc}{2\lambda_B} = W_0 + K_A$ ... $(iii)$
From $(ii)$,we have $W_0 = \frac{hc}{\lambda_B} - K_B$. Substitute this into $(iii)$:
$\frac{hc}{2\lambda_B} = (\frac{hc}{\lambda_B} - K_B) + K_A$
Rearranging the terms:
$K_A = K_B - \frac{hc}{2\lambda_B}$
Since $\frac{hc}{\lambda_B} = W_0 + K_B$,then $\frac{hc}{2\lambda_B} = \frac{W_0 + K_B}{2}$.
Substituting this back:
$K_A = K_B - \frac{W_0 + K_B}{2} = \frac{2K_B - W_0 - K_B}{2} = \frac{K_B - W_0}{2} = \frac{K_B}{2} - \frac{W_0}{2}$.
Since $W_0 > 0$,it follows that $K_A < \frac{K_B}{2}$.

(A) The work function $\Phi_0$ is given by the formula $\Phi_0 = \frac{hc}{\lambda_0}$.
Using the shortcut formula $\Phi_0 \approx \frac{12400}{\lambda_0 (\text{in } \mathring{A})} \text{ eV}$.
Substituting the given threshold wavelength $\lambda_0 = 6500 \mathring{A}$:
$\Phi_0 = \frac{12400}{6500} \text{ eV}$.
$\Phi_0 \approx 1.907 \text{ eV}$.
Rounding to the nearest integer,we get $\Phi_0 \approx 2 \text{ eV}$.

(A) The photoelectric effect occurs when the incident radiation has a frequency greater than the threshold frequency of the metal, or equivalently, a wavelength shorter than the threshold wavelength.
Since ultraviolet $(UV)$ rays do not cause the photoelectric effect, the incident radiation must have a higher energy (shorter wavelength) than $UV$ rays.
Among the given options, $X$-rays have a shorter wavelength $(\lambda_{X-ray} < \lambda_{UV-ray})$ and thus higher energy than $UV$ rays.
Therefore, $X$-rays will cause the photoelectric effect.

(A) According to Einstein's photoelectric equation,the energy of the incident photon is equal to the sum of the work function and the maximum kinetic energy of the emitted photoelectrons.
$E = \Phi + K_{\max}$
Here,$E = h\nu = h(4\nu_0)$ and the work function $\Phi = h\nu_0$.
Substituting these values into the equation:
$h(4\nu_0) = h\nu_0 + K_{\max}$
$K_{\max} = 4h\nu_0 - h\nu_0$
$K_{\max} = 3h\nu_0$.

(A) Einstein explained the photoelectric effect by proposing that light consists of discrete packets of energy called photons. The energy of each photon is given by the equation $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency of the radiation. This equation forms the basis of Einstein's photoelectric equation,$K_{max} = h\nu - \Phi_0$. Therefore,the correct option is $A$.

(C) The work function $W_0$ is given by the relation $W_0 = \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.
Using the approximation $hc \approx 12375 \ eV \cdot \mathring A$,we have:
$\lambda_0 = \frac{12375}{W_0} \mathring A$
Substituting $W_0 = 2.3 \ eV$:
$\lambda_0 = \frac{12375}{2.3} \approx 5380.43 \mathring A$
Rounding to the nearest given option,we get $5380 \mathring A$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = hf - W_0$,where $W_0$ is the work function.
For the first photo-cathode: $hf_1 = W_0 + \frac{1}{2}mv_1^2$ ... $(i)$
For the second photo-cathode: $hf_2 = W_0 + \frac{1}{2}mv_2^2$ ... $(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$h(f_1 - f_2) = \frac{1}{2}m(v_1^2 - v_2^2)$
Rearranging the terms to solve for the velocity difference:
$v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$

(B) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by:
$\frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) = V_0$
For the first case:
$\frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) = 4.8 \quad ...(i)$
For the second case,where wavelength is $2\lambda$:
$\frac{hc}{e} \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) = 1.6 \quad ...(ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}} = \frac{4.8}{1.6} = 3$
$\frac{1}{\lambda} - \frac{1}{\lambda_0} = 3 \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right)$
$\frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{3}{2\lambda} - \frac{3}{\lambda_0}$
$\frac{3}{\lambda_0} - \frac{1}{\lambda_0} = \frac{3}{2\lambda} - \frac{1}{\lambda} = \frac{1}{2\lambda}$
$\frac{2}{\lambda_0} = \frac{1}{2\lambda}$
$\lambda_0 = 4\lambda$

(C) The photoelectric effect occurs only when the energy of the incident photon is greater than or equal to the work function of the metal.
The energy of an incident photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
The work function $\phi_0$ is related to the threshold frequency $\nu_0$ by the equation $\phi_0 = h\nu_0$.
For the emission of photoelectrons,the condition is $E \ge \phi_0$.
Substituting the values,we get $h\nu \ge h\nu_0$,which simplifies to $\nu \ge \nu_0$.
Therefore,the necessary condition for the emission of photoelectrons is $\nu \ge \nu_0$.

(D) According to Einstein's photoelectric equation,$E = W_0 + K_{\max}$,where $E = \frac{hc}{\lambda}$ is the energy of the incident photon,$W_0$ is the work function,and $K_{\max}$ is the maximum kinetic energy.
Using the relation $E \approx \frac{12400}{\lambda (\text{in } \mathring A)} \ eV$:
For $\lambda_1 = 1824 \ \mathring A$,$E_1 = \frac{12400}{1824} \approx 6.797 \ eV \approx 6.8 \ eV$.
Given $K_{\max, 1} = 5.3 \ eV$,then $W_0 = E_1 - K_{\max, 1} = 6.8 - 5.3 = 1.5 \ eV$.
For $\lambda_2 = 1216 \ \mathring A$,$E_2 = \frac{12400}{1216} \approx 10.197 \ eV \approx 10.2 \ eV$.
Given $K_{\max, 2} = 8.7 \ eV$,then $W_0 = E_2 - K_{\max, 2} = 10.2 - 8.7 = 1.5 \ eV$.
Thus,the work function of the metal surface is $1.5 \ eV$.

(B) The work function $\Phi_0$ is given by $\Phi_0 = \frac{hc}{\lambda_0}$.
Here,$\Phi_0 = 6.825 \; eV$.
Using the relation $\lambda_0 (\text{in } \mathring A) = \frac{12400}{\Phi_0 (\text{in } eV)}$ or the more precise constant $\lambda_0 = \frac{12375}{\Phi_0}$.
$\lambda_0 = \frac{12375}{6.825} \approx 1813 \; \mathring A$.
Rounding to the nearest given option,we get $1800 \; \mathring A$.

(C) The work function $W_0$ is given by $W_0 = h\nu_0$.
Given $h = 6.6 \times 10^{-34} Js$ and $\nu_0 = 1.6 \times 10^{15} Hz$.
$W_0 = (6.6 \times 10^{-34}) \times (1.6 \times 10^{15}) = 10.56 \times 10^{-19} J$.
To convert this into $eV$,divide by the charge of an electron $(1.6 \times 10^{-19} C)$:
$W_0 = \frac{10.56 \times 10^{-19}}{1.6 \times 10^{-19}} eV = 6.6 eV$.
According to Einstein's photoelectric equation,$K_{max} = E - W_0$.
Given incident energy $E = 8 eV$.
$K_{max} = 8 eV - 6.6 eV = 1.4 eV$.

(D) The photoelectric effect is explained by the $Quantum$ theory of light.
According to this theory,light consists of discrete packets of energy called photons.
When a photon with energy $E = h\nu$ is incident on a metal surface,it transfers its energy to an electron.
If the energy of the incident photon is greater than the work function $(\Phi)$ of the metal,the electron overcomes the binding force and is emitted from the surface.
This phenomenon cannot be explained by the wave theory of light because the wave theory predicts that the emission of electrons should depend on the intensity of light,whereas the photoelectric effect depends on the frequency of the incident light.

(B) According to Einstein's photoelectric equation,$K.E._{max} = h\nu - \Phi$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\Phi$ is the work function of the metal.
Since $\Phi$ is a constant for a given metal,the maximum kinetic energy of the emitted photoelectrons depends directly on the frequency of the incident light $(\nu)$.
Increasing the intensity of light increases the number of photoelectrons emitted per second,but it does not affect the kinetic energy of individual electrons.
Therefore,the correct option is $(B)$.

(B) The work function $W_0$ is related to the threshold wavelength $\lambda_0$ by the formula $W_0 = \frac{hc}{\lambda_0}$.
Using the approximation $hc \approx 12375 \, eV \cdot \mathring{A}$,we have:
$W_0 = \frac{12375}{\lambda_0} \, eV$.
Given $\lambda_0 = 5420 \, \mathring{A}$,
$W_0 = \frac{12375}{5420} \approx 2.28 \, eV$.
Therefore,the correct option is $B$.

(C) The energy $E$ of the incident photon is given by the formula $E = \frac{hc}{\lambda}$.
Using the approximation $hc \approx 12400 \text{ eV} \cdot \mathring{A}$,we calculate the energy as:
$E = \frac{12400}{4100} \approx 3.02 \text{ eV}$.
Photoelectric emission occurs if the energy of the incident photon is greater than the work function $(\Phi)$ of the metal.
For metal $A$: $\Phi_A = 1.92 \text{ eV} < 3.02 \text{ eV}$ (Emission occurs).
For metal $B$: $\Phi_B = 2.0 \text{ eV} < 3.02 \text{ eV}$ (Emission occurs).
For metal $C$: $\Phi_C = 5.0 \text{ eV} > 3.02 \text{ eV}$ (No emission).
Therefore,only metals $A$ and $B$ will emit photoelectrons.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = E - W_0$,where $E$ is the energy of the incident photon and $W_0$ is the work function.
For the first case: $E_1 = 2h\nu_0$ and $W_0 = h\nu_0$.
$\frac{1}{2}mv_1^2 = 2h\nu_0 - h\nu_0 = h\nu_0$ ... $(i)$
For the second case: $E_2 = 5h\nu_0$ and $W_0 = h\nu_0$.
$\frac{1}{2}mv_2^2 = 5h\nu_0 - h\nu_0 = 4h\nu_0$ ... $(ii)$
Dividing equation $(ii)$ by $(i)$:
$\frac{\frac{1}{2}mv_2^2}{\frac{1}{2}mv_1^2} = \frac{4h\nu_0}{h\nu_0}$
$\left(\frac{v_2}{v_1}\right)^2 = 4$
$\frac{v_2}{v_1} = 2$
Given $v_1 = 4 \times 10^6 \, m/s$,we find:
$v_2 = 2 \times v_1 = 2 \times (4 \times 10^6 \, m/s) = 8 \times 10^6 \, m/s$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of the emitted photoelectrons is given by:
$K_{max} = E - \Phi$
Where $E$ is the energy of the incident photon and $\Phi$ is the work function of the metal.
Given: $E = 1.8 \ eV$ and $\Phi = 1.2 \ eV$.
$K_{max} = 1.8 \ eV - 1.2 \ eV = 0.6 \ eV$.
The stopping potential $(V_s)$ is related to the maximum kinetic energy by the relation $K_{max} = e V_s$.
Therefore,the stopping potential is $0.6 \ V$,which corresponds to a value of $0.6 \ eV$ in terms of energy units.

(A) The behavior of an incident photon in the photoelectric effect depends on the frequency of the incident light relative to the threshold frequency of the metal surface.
Case $I$: If the frequency of the incident light is less than the threshold frequency $(\nu < \nu_0)$, no photoelectric emission occurs. In this case, the photon is scattered or reflected by the surface, and it comes out without any change in its frequency.
Case $II$: If the frequency of the incident light is greater than or equal to the threshold frequency $(\nu \geq \nu_0)$, the photon is absorbed by an electron. The energy of the photon $(E = h\nu)$ is used to overcome the work function $(\Phi)$ of the metal and provide kinetic energy to the emitted electron. Consequently, the photon completely disappears.

(A) The maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by Einstein's photoelectric equation:
$K_{max} = E - \Phi$
where $E$ is the energy of the incident photon and $\Phi$ is the work function of the metal.
The work function $\Phi$ is given by $\Phi = h\nu_0$,where $\nu_0$ is the threshold frequency.
Given $h = 6 \times 10^{-34} \ J \cdot s$ and $\nu_0 = 1.6 \times 10^{15} \ Hz$:
$\Phi = (6 \times 10^{-34} \ J \cdot s) \times (1.6 \times 10^{15} \ Hz) = 9.6 \times 10^{-19} \ J$.
To convert the work function into $eV$,divide by $1.6 \times 10^{-19} \ J/eV$:
$\Phi = \frac{9.6 \times 10^{-19}}{1.6 \times 10^{-19}} \ eV = 6 \ eV$.
Now,calculate the maximum kinetic energy:
$K_{max} = 8 \ eV - 6 \ eV = 2 \ eV$.

(D) Using Einstein's photoelectric equation: $K_{max} = E - \Phi$,where $\Phi$ is the work function.
For metal $A$: $T_A = 4.25 - \Phi_A$ ... $(i)$
For metal $B$: $T_B = T_A - 1.50 = 4.70 - \Phi_B$ ... (ii)
From $(i)$,$\Phi_A = 4.25 - T_A$. From (ii),$\Phi_B = 4.70 - (T_A - 1.50) = 6.20 - T_A$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mK}}$,so $\lambda \propto \frac{1}{\sqrt{K}}$.
Given $\lambda_B = 2\lambda_A$,we have $\frac{\lambda_B}{\lambda_A} = \sqrt{\frac{T_A}{T_B}} = 2$.
Squaring both sides: $\frac{T_A}{T_A - 1.50} = 4$.
$T_A = 4T_A - 6.00 \Rightarrow 3T_A = 6.00 \Rightarrow T_A = 2.00\, eV$.
Substituting $T_A = 2.00\, eV$ into $(i)$: $\Phi_A = 4.25 - 2.00 = 2.25\, eV$.
Substituting $T_A = 2.00\, eV$ into (ii): $T_B = 2.00 - 1.50 = 0.50\, eV$. Then $\Phi_B = 4.70 - 0.50 = 4.20\, eV$.
Thus,all statements are correct.

(D) The intensity of the image formed by a lens is proportional to the power collected by the lens and inversely proportional to the area of the image.
The power collected by the lens is proportional to the area of the lens aperture,which remains constant since the diameter is the same.
The area of the image $A$ is proportional to the square of the focal length $(A \propto f^2)$.
Therefore,the intensity $E$ is proportional to $1/f^2$.
Initially,$I \propto 1/f_1^2 = 1/(30)^2$.
Finally,$I' \propto 1/f_2^2 = 1/(15)^2$.
Taking the ratio: $I'/I = (f_1/f_2)^2 = (30/15)^2 = 2^2 = 4$.
Thus,the new photoelectric current is $I' = 4I$.

(C) According to Einstein's photoelectric equation:
$E = \phi + K_{max}$
$\frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2$
Rearranging for $v$:
$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi$
$\frac{1}{2}mv^2 = \frac{hc - \lambda \phi}{\lambda}$
$v^2 = \frac{2(hc - \lambda \phi)}{m \lambda}$
$v = [\frac{2(hc - \lambda \phi)}{m \lambda}]^{1/2}$

(B) The cut-off voltage (stopping potential) depends only on the frequency of the incident light,not on its intensity. Therefore,it remains $0.6 \ V$.
The intensity of light from a point source follows the inverse square law,$I \propto \frac{1}{d^2}$.
When the distance increases from $0.2 \ m$ to $0.6 \ m$,the distance becomes $3$ times $(d' = 3d)$.
Thus,the intensity becomes $I' = \frac{I}{3^2} = \frac{I}{9}$.
Since the saturation current is directly proportional to the intensity of incident light,the new saturation current will be $I'_{sat} = \frac{18 \ mA}{9} = 2 \ mA$.
Comparing this with the given options,option $(b)$ is correct regarding the stopping potential.

(D) According to Einstein's photoelectric equation: $K_{\max} = \frac{hc}{\lambda} - W_0 = \frac{1}{2}mv^2$,where $W_0 = \frac{hc}{\lambda_0}$ is the work function.
Thus,$v = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)} = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right)} \dots (i)$
When the wavelength is changed to $\lambda' = \frac{3\lambda}{4}$,the new velocity $v'$ is:
$v' = \sqrt{\frac{2hc}{m} \left( \frac{1}{3\lambda/4} - \frac{1}{\lambda_0} \right)} = \sqrt{\frac{2hc}{m} \left( \frac{4\lambda_0 - 3\lambda}{3\lambda \lambda_0} \right)} \dots (ii)$
Dividing $(ii)$ by $(i)$:
$\frac{v'}{v} = \sqrt{\frac{4\lambda_0 - 3\lambda}{3\lambda \lambda_0} \cdot \frac{\lambda \lambda_0}{\lambda_0 - \lambda}} = \sqrt{\frac{4}{3} \cdot \frac{\lambda_0 - 0.75\lambda}{\lambda_0 - \lambda}}$
Since $\lambda_0 > \lambda$,it follows that $(\lambda_0 - 0.75\lambda) > (\lambda_0 - \lambda)$.
Therefore,$\frac{\lambda_0 - 0.75\lambda}{\lambda_0 - \lambda} > 1$.
This implies $\frac{v'}{v} > \sqrt{\frac{4}{3}}$,or $v' > v(4/3)^{1/2}$.

(C) The intensity of light $I$ is given by $I = \frac{n E}{A t}$,where $n$ is the number of photons,$E = \frac{hc}{\lambda}$ is the energy of one photon,$A$ is the area,and $t$ is time.
The number of incident photons per second is $N = \frac{I A}{E} = \frac{I A \lambda}{hc}$.
Given:
$I = 1.0 \ W/m^2$
$A = 1.0 \ cm^2 = 1.0 \times 10^{-4} \ m^2$
$\lambda = 300 \ nm = 300 \times 10^{-9} \ m$
$h = 6.63 \times 10^{-34} \ J \cdot s$
$c = 3 \times 10^8 \ m/s$
$N = \frac{1.0 \times 10^{-4} \times 300 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8} = \frac{3 \times 10^{-11}}{19.89 \times 10^{-26}} \approx 1.508 \times 10^{14} \text{ photons/sec}$.
Since $1\%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted per second is:
$N_e = 0.01 \times N = 0.01 \times 1.508 \times 10^{14} = 1.508 \times 10^{12} \text{ per sec}$.
Thus,the correct option is $C$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy ${K_{\max }}$ is given by ${K_{\max }} = h\nu - h{\nu _0}$,where ${\nu _0}$ is the threshold frequency.
For frequency ${\nu _1}$,${K_1} = h({\nu _1} - {\nu _0})$.
For frequency ${\nu _2}$,${K_2} = h({\nu _2} - {\nu _0})$.
Given the ratio of maximum kinetic energies is ${K_1}:{K_2} = 1:k$,we have $\frac{{{K_1}}}{{{K_2}}} = \frac{1}{k}$.
Substituting the expressions,we get $\frac{{h({\nu _1} - {\nu _0})}}{{h({\nu _2} - {\nu _0})}} = \frac{1}{k}$.
$k({\nu _1} - {\nu _0}) = {\nu _2} - {\nu _0}$.
$k{\nu _1} - k{\nu _0} = {\nu _2} - {\nu _0}$.
$k{\nu _1} - {\nu _2} = k{\nu _0} - {\nu _0} = {\nu _0}(k - 1)$.
Therefore,the threshold frequency is ${\nu _0} = \frac{{k{\nu _1} - {\nu _2}}}{{k - 1}}$.

(B) The energy of the emitted photons from a hydrogen discharge tube corresponds to the ionization energy of hydrogen,which is $E = 13.6 \ eV$.
According to Einstein's photoelectric equation,$E = W_0 + K_{max}$,where $K_{max} = e|V_0|$ and $V_0$ is the stopping potential.
Substituting the given values: $13.6 \ eV = 4.2 \ eV + e|V_0|$.
$e|V_0| = 13.6 \ eV - 4.2 \ eV = 9.4 \ eV$.
Therefore,the magnitude of the stopping potential is $|V_0| = 9.4 \ V$.
To reduce the photo-current to zero,the anode must be at a negative potential relative to the cathode.
Thus,the required voltage is $-9.4 \ V$.

(A) The threshold wavelength $\lambda_0$ is given by the formula $\lambda_0 = \frac{hc}{W_0}$.
Using the approximation $\lambda_0 (\text{in } \mathring{A}) \approx \frac{12400}{W_0 (\text{in } eV)}$,we calculate the threshold wavelength for both metals.
For Lithium $(Li)$: $(\lambda_0)_{Li} = \frac{12400}{2.3} \approx 5391 \ \mathring{A}$.
For Copper $(Cu)$: $(\lambda_0)_{Cu} = \frac{12400}{4.0} = 3100 \ \mathring{A}$.
The visible light spectrum ranges approximately from $3800 \ \mathring{A}$ to $7500 \ \mathring{A}$.
Since $(\lambda_0)_{Li} = 5391 \ \mathring{A}$ falls within the visible region,Lithium can emit photoelectrons when exposed to visible light.
Since $(\lambda_0)_{Cu} = 3100 \ \mathring{A}$ falls in the ultraviolet region,Copper cannot emit photoelectrons with visible light.
Therefore,Lithium is the correct metal for a photoelectric cell working with visible light.

(C) $1$. Calculate the total number of incident photons $(n)$:
$n = \frac{E \lambda}{hc} = \frac{1 \times 10^{-7} \times 200 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8} \approx 1.005 \times 10^{11} \approx 10^{11}$ photons.
$2$. Calculate the number of ejected electrons $(N_e)$:
Given that $1$ out of $10^3$ photons ejects an electron,$N_e = \frac{n}{10^3} = \frac{10^{11}}{10^3} = 10^8$ electrons.
$3$. Calculate the total charge $(q)$ on the sphere:
$q = N_e \times e = 10^8 \times 1.6 \times 10^{-19} \ C = 1.6 \times 10^{-11} \ C$.
$4$. Calculate the potential $(V)$ on the sphere:
$V = \frac{1}{4\pi \epsilon_0} \frac{q}{r} = (9 \times 10^9) \times \frac{1.6 \times 10^{-11}}{4.8 \times 10^{-2}} = \frac{14.4 \times 10^{-2}}{4.8 \times 10^{-2}} = 3 \ V$.

(D) According to Einstein's photoelectric equation:
$K_{\max} = h\nu - \phi_0$
where $K_{\max}$ is the maximum kinetic energy of the ejected photoelectrons,$h$ is Planck's constant,$\nu$ is the frequency of incident radiation,and $\phi_0 = h\nu_0$ is the work function of the metal.
Comparing this with the equation of a straight line $y = mx + c$:
Here,$y = K_{\max}$,$x = \nu$,$m = h$ (slope),and $c = -\phi_0$ (y-intercept).
Since the slope $h$ is positive and the intercept $-\phi_0$ is negative,the graph is a straight line that starts from a threshold frequency $\nu_0$ on the x-axis and has a positive slope. This corresponds to the graph where the line intersects the x-axis at $\nu_0$ and increases linearly for $\nu > \nu_0$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $E_k$ of the emitted photoelectrons is given by:
$E_k = h\nu - \Phi$
where $h$ is Planck's constant,$\nu$ is the frequency of incident photons,and $\Phi$ is the work function of the metal.
Comparing this equation with the equation of a straight line $y = mx + c$,where $y = E_k$,$x = \nu$,$m$ is the slope,and $c$ is the y-intercept:
$E_k = h\nu + (-\Phi)$
Here,the slope $m = h$.
Therefore,the slope of the curve represents Planck's constant.

(B) According to Einstein's photoelectric equation:
$K_{\max} = h\nu - h\nu_0$
Since $K_{\max} = eV_0$,where $V_0$ is the stopping potential and $e$ is the charge of an electron:
$eV_0 = h\nu - h\nu_0$
Dividing by $e$:
$V_0 = (\frac{h}{e})\nu - (\frac{h\nu_0}{e})$
Comparing this with the equation of a straight line $y = mx + c$,where $y = V_0$ and $x = \nu$:
The slope $m = \frac{h}{e}$.
Therefore,Planck's constant $h = m \times e$,which is the product of the slope of the line and the charge on the electron.