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

(D) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function of the metal.
For wavelength $\lambda$,the stopping potential is $eV_1 = \frac{hc}{\lambda} - \phi$ --- $(1)$.
For wavelength $\frac{\lambda}{2}$,the stopping potential is $eV_2 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$ --- $(2)$.
From equation $(1)$,we have $\frac{hc}{\lambda} = eV_1 + \phi$.
Substituting this into equation $(2)$:
$eV_2 = 2(eV_1 + \phi) - \phi$
$eV_2 = 2eV_1 + 2\phi - \phi$
$eV_2 = 2eV_1 + \phi$
Since the work function $\phi > 0$,it follows that $eV_2 > 2eV_1$,which implies $V_2 > 2V_1$.

(A) The maximum kinetic energy is given by Einstein's photoelectric equation:
$K_{\max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$
where $\lambda_0 = 3250 \times 10^{-10} \, m$ is the threshold wavelength and $\lambda = 2536 \times 10^{-10} \, m$ is the incident wavelength.
Given $hc = (4.14 \times 10^{-15} \, eV \cdot s) \times (3 \times 10^8 \, m/s) = 12420 \, eV \cdot \mathring{A}$.
Converting wavelengths to $\mathring{A}$s: $\lambda_0 = 3250 \, \mathring{A}$ and $\lambda = 2536 \, \mathring{A}$.
$K_{\max} = 12420 \left( \frac{1}{2536} - \frac{1}{3250} \right) \, eV = 12420 \left( \frac{3250 - 2536}{2536 \times 3250} \right) \, eV \approx 1.076 \, eV$.
Converting $K_{\max}$ to Joules: $K_{\max} = 1.076 \times 1.6 \times 10^{-19} \, J \approx 1.72 \times 10^{-19} \, J$.
Using $K_{\max} = \frac{1}{2}mv^2$ with $m = 9.1 \times 10^{-31} \, kg$:
$v^2 = \frac{2 \times 1.72 \times 10^{-19}}{9.1 \times 10^{-31}} \approx 0.378 \times 10^{12} \, m^2/s^2$.
$v \approx 0.615 \times 10^6 \, m/s \approx 6 \times 10^5 \, m/s$.

(A) According to Einstein's photoelectric equation,$K_{max} = h\nu - W_0 = \frac{1}{2}mv^2$,where $W_0 = h\nu_0$.
For frequency $2\nu_0$:
$h(2\nu_0) = h\nu_0 + \frac{1}{2}mv_1^2$
$h\nu_0 = \frac{1}{2}mv_1^2$ ..... $(i)$
For frequency $5\nu_0$:
$h(5\nu_0) = h\nu_0 + \frac{1}{2}mv_2^2$
$4h\nu_0 = \frac{1}{2}mv_2^2$ ..... $(ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{h\nu_0}{4h\nu_0} = \frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2}$
$\frac{1}{4} = \frac{v_1^2}{v_2^2}$
Taking the square root on both sides:
$\frac{v_1}{v_2} = \frac{1}{2}$
Thus,the ratio is $1:2$.

(C) Given,stopping potential $V_{0} = 10 \ V$ and work function $W = 2.75 \ eV$.
According to Einstein's photoelectric equation,the energy of the incident photon is $E = h\nu = eV_{0} + W$.
Substituting the values,$E = 10 \ eV + 2.75 \ eV = 12.75 \ eV$ ..... $(i)$.
For a hydrogen atom,the energy of a photon emitted during a transition from state $n$ to the ground state $(n=1)$ is given by $h\nu = E_{n} - E_{1}$.
Since $E_{n} = -\frac{13.6}{n^{2}} \ eV$,we have $h\nu = -\frac{13.6}{n^{2}} - (-13.6) = 13.6 \left(1 - \frac{1}{n^{2}}\right) \ eV$.
Equating this to the energy from $(i)$,$13.6 \left(1 - \frac{1}{n^{2}}\right) = 12.75$.
$1 - \frac{1}{n^{2}} = \frac{12.75}{13.6} = 0.9375$.
$\frac{1}{n^{2}} = 1 - 0.9375 = 0.0625$.
$n^{2} = \frac{1}{0.0625} = 16$.
Therefore,$n = 4$.

(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
For photoelectric emission to occur, the energy of the incident photon must be greater than or equal to the work function of the metal $(E \ge \phi)$.
This implies that the wavelength of the incident light must be less than or equal to the threshold wavelength $(\lambda \le \lambda_0)$.
Given that green light causes emission but yellow light does not, the threshold wavelength $\lambda_0$ lies between the wavelengths of green and yellow light $(\lambda_{\text{green}} < \lambda_0 < \lambda_{\text{yellow}})$.
Since the wavelength of light increases in the order $V < I < B < G < Y < O < R$, any color with a wavelength shorter than green light will have higher energy and will cause photoelectric emission.
Among the given options, Indigo has a shorter wavelength than green light.
Therefore, Indigo can produce the emission of photo-electrons.

(C) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by $eV_0 = h\nu - W_0$,where $W_0$ is the work function.
This can be rewritten as $V_0 = \frac{h}{e}\nu - \frac{W_0}{e}$.
The graph between $V_0$ and $\nu$ is a straight line with a slope of $\frac{h}{e}$.
The $x$-intercept of this graph is the threshold frequency $\nu_0$,where $V_0 = 0$,so $h\nu_0 = W_0$.
From the graph,the threshold frequencies are in the order: $(\nu_0)_{(iv)} > (\nu_0)_{(iii)} > (\nu_0)_{(ii)} > (\nu_0)_{(i)}$.
Since the work function $W_0 = h\nu_0$,the work functions follow the same order: $(W_0)_{(iv)} > (W_0)_{(iii)} > (W_0)_{(ii)} > (W_0)_{(i)}$.
Therefore,the correct ranking is $(iv) > (iii) > (ii) > (i)$.

(C) According to Einstein's photoelectric equation,the energy of the incident photon is equal to the sum of the work function $(W_0)$ and the maximum kinetic energy $(K_{max})$ of the emitted photoelectrons: $E = W_0 + K_{max}$.
For wavelength $\lambda_1$: $\frac{hc}{\lambda_1} = W_0 + E_1$ --- $(1)$
For wavelength $\lambda_2$: $\frac{hc}{\lambda_2} = W_0 + E_2$ --- $(2)$
From $(1)$,$hc = W_0 \lambda_1 + E_1 \lambda_1$. From $(2)$,$hc = W_0 \lambda_2 + E_2 \lambda_2$.
Equating the two expressions for $hc$: $W_0 \lambda_1 + E_1 \lambda_1 = W_0 \lambda_2 + E_2 \lambda_2$.
Rearranging to solve for $W_0$: $W_0 (\lambda_1 - \lambda_2) = E_2 \lambda_2 - E_1 \lambda_1$.
$W_0 = \frac{E_2 \lambda_2 - E_1 \lambda_1}{\lambda_1 - \lambda_2} = \frac{E_1 \lambda_1 - E_2 \lambda_2}{\lambda_2 - \lambda_1}$.

(B) Given: Maximum velocity $v_{max} = 4 \times 10^8 \, cm/s = 4 \times 10^6 \, m/s$.
Mass of electron $m = 9 \times 10^{-31} \, kg$.
Charge of electron $e = 1.6 \times 10^{-19} \, C$.
The maximum kinetic energy is given by $K_{max} = \frac{1}{2} m v_{max}^2$.
$K_{max} = \frac{1}{2} \times (9 \times 10^{-31} \, kg) \times (4 \times 10^6 \, m/s)^2$.
$K_{max} = 0.5 \times 9 \times 10^{-31} \times 16 \times 10^{12} \, J$.
$K_{max} = 72 \times 10^{-19} \, J$.
To convert this into electron-volts $(eV)$,divide by $e = 1.6 \times 10^{-19} \, C$:
$K_{max} = \frac{72 \times 10^{-19}}{1.6 \times 10^{-19}} \, eV = 45 \, eV$.
The stopping potential $V_0$ is defined as $K_{max} = e V_0$,so $V_0 = 45 \, V$.

(B) According to Einstein's photoelectric equation: $hf = W_0 + \frac{1}{2}mv_{\max}^2$,where $W_0$ is the work function of the metal.
Thus,$v_{\max} = \sqrt{\frac{2(hf - W_0)}{m}} = V$.
When the frequency becomes $4f$,the new maximum velocity $V'$ is given by:
$V' = \sqrt{\frac{2(h(4f) - W_0)}{m}}$.
We can rewrite this as: $V' = \sqrt{\frac{2(4hf - W_0)}{m}}$.
Since $4hf - W_0 > 4(hf - W_0)$,we have $V' > \sqrt{\frac{2(4(hf - W_0))}{m}} = 2\sqrt{\frac{2(hf - W_0)}{m}} = 2V$.
Therefore,$V' > 2V$.

(D) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \phi + K$ $...(1)$
When the wavelength is changed to $\lambda' = \frac{3\lambda}{4}$,the new kinetic energy $K'$ is given by:
$\frac{hc}{3\lambda/4} = \phi + K'$ $...(2)$
$\frac{4hc}{3\lambda} = \phi + K'$ $...(2)$
From equation $(1)$,we have $\phi = \frac{hc}{\lambda} - K$. Substituting this into equation $(2)$:
$\frac{4}{3} \left( \frac{hc}{\lambda} \right) = \left( \frac{hc}{\lambda} - K \right) + K'$
$\frac{4}{3} \left( \frac{hc}{\lambda} \right) - \frac{hc}{\lambda} = K' - K$
$\frac{1}{3} \left( \frac{hc}{\lambda} \right) = K' - K$
Since $\frac{hc}{\lambda} = \phi + K$,we substitute this back:
$K' = K + \frac{1}{3} (\phi + K) = K + \frac{\phi}{3} + \frac{K}{3} = \frac{4K}{3} + \frac{\phi}{3}$
Since the work function $\phi > 0$,it follows that $K' > \frac{4K}{3}$.

(C) The maximum kinetic energy for the photoelectrons is given by Einstein's photoelectric equation: $K_{\max} = h\nu - \phi$,where $\nu$ is the frequency of incident light and $\phi$ is the work function of the metal.
Since $K_{\max} = eV_0$,we have $eV_0 = h\nu - \phi$ --- $(1)$
When the frequency is doubled $(\nu' = 2\nu)$,the new stopping potential $V_0'$ is given by:
$eV_0' = h(2\nu) - \phi = 2h\nu - \phi$
We can rewrite this as:
$eV_0' = 2(h\nu - \frac{\phi}{2})$ --- $(2)$
Comparing $(1)$ and $(2)$,since $h\nu - \frac{\phi}{2} > h\nu - \phi$,it follows that $eV_0' > 2eV_0$. Therefore,the stopping potential becomes more than double.

(A) According to the Einstein's photoelectric equation, the interaction between a photon and an electron is a one-to-one process.
When a single photon of energy $E = h\nu$ strikes a metal surface, it is completely absorbed by a single electron.
If the energy of the photon is greater than the work function $(\Phi)$ of the metal, this single electron gains enough kinetic energy to overcome the surface barrier and is emitted as a photoelectron.
Therefore, one photon can eject at most one electron. If the photon energy is less than the work function, no electron is emitted.
Thus, the number of electrons emitted depends on the number of incident photons and their energy relative to the work function.

(B) The maximum kinetic energy $(K_{max})$ of photoelectrons is given by Einstein's photoelectric equation: $K_{max} = h\nu - \phi$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\phi$ is the work function of the material.
When the frequency $\nu$ is doubled,the term $h\nu$ increases,which leads to an increase in $K_{max}$. However,$K_{max}$ does not simply double because of the subtraction of the constant work function $\phi$.
The photoelectric current is directly proportional to the intensity of the incident light. When the intensity is doubled,the number of incident photons per unit time doubles,which in turn doubles the number of photoelectrons emitted per unit time,thereby doubling the photoelectric current.
Therefore,the correct description is that the kinetic energy increases (but not necessarily by a factor of two) and the photoelectric current increases by a factor of two. Among the given options,option $B$ is the most appropriate description of the physical phenomena.

(D) The work function of the cathode surface is $\phi = 4 \ eV$.
In a hydrogen discharge tube,the maximum energy of a photon emitted during the transition of an electron from higher energy levels to the ground state $(n=1)$ is given by the Rydberg formula,where the maximum energy corresponds to the transition from $n = \infty$ to $n = 1$,which is $13.6 \ eV$.
According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of the emitted photoelectrons is given by $K_{max} = E_{photon} - \phi$.
Substituting the maximum photon energy,$K_{max} = 13.6 \ eV - 4 \ eV = 9.6 \ eV$.
To make the photocurrent zero,the stopping potential $(V_s)$ must be applied such that $e|V_s| = K_{max}$.
Therefore,$|V_s| = 9.6 \ V$.
Since the stopping potential must be negative with respect to the cathode to stop the electrons,the anode voltage must be at least $-9.6 \ V$. Among the given options,$-10 \ V$ is the only value that is more negative than $-9.6 \ V$,which will effectively stop all photoelectrons.

(C) In the photoelectric effect, the stopping potential $(V_s)$ depends only on the frequency $(\nu)$ of the incident radiation and the work function $(\Phi)$ of the metal surface, according to Einstein's photoelectric equation: $eV_s = h\nu - \Phi$.
Moving the point source farther away decreases the intensity of the incident light, which reduces the number of photoelectrons emitted per unit time (photoelectric current).
However, the intensity does not affect the maximum kinetic energy of the emitted photoelectrons, as the frequency of the light remains unchanged.
Since the stopping potential is determined solely by the maximum kinetic energy, it remains constant regardless of the distance of the source.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - W$,where $W$ is the work function of the metal.
For wavelength $\lambda_1$,$K_1 = \frac{hc}{\lambda_1} - W$ $(i)$
For wavelength $\lambda_2$,$K_2 = \frac{hc}{\lambda_2} - W$ $(ii)$
Given $\lambda_1 = 2\lambda_2$,substitute this into equation $(i)$:
$K_1 = \frac{hc}{2\lambda_2} - W$
$K_1 = \frac{1}{2} \left( \frac{hc}{\lambda_2} \right) - W$
From equation $(ii)$,we know $\frac{hc}{\lambda_2} = K_2 + W$.
Substituting this into the expression for $K_1$:
$K_1 = \frac{1}{2} (K_2 + W) - W$
$K_1 = \frac{K_2}{2} + \frac{W}{2} - W$
$K_1 = \frac{K_2}{2} - \frac{W}{2}$
Since $W > 0$,it follows that $K_1 < \frac{K_2}{2}$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by:
$K_{max} = hf - \phi$
where $h$ is Planck's constant,$f$ is the frequency of incident light,and $\phi$ is the work function of the metal.
In the photoelectric effect,the stopping potential $V$ is related to the maximum kinetic energy by the relation $K_{max} = eV$,where $e$ is the charge of an electron.
Thus,$eV = hf - \phi$,or $V = \frac{h}{e}f - \frac{\phi}{e}$.
From the graph,when $f = f_0$ (threshold frequency),the stopping potential $V = 0$. Therefore,$0 = \frac{h}{e}f_0 - \frac{\phi}{e}$,which implies $\phi = hf_0$.
Substituting this into the expression for $K_{max}$ at frequency $f_1$:
$K_{max} = hf_1 - hf_0 = h(f_1 - f_0)$.
Therefore,the correct option is $C$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E._{max})$ of emitted photoelectrons is given by:
$K.E._{max} = E - W$,where $E$ is the incident photon energy and $W$ is the work function.
For the first case,$E_1 = 2W$:
$\frac{1}{2} m v_1^2 = 2W - W = W$ --- $(1)$
For the second case,$E_2 = 5W$:
$\frac{1}{2} m v_2^2 = 5W - W = 4W$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{v_1^2}{v_2^2} = \frac{W}{4W} = \frac{1}{4}$
Taking the square root on both sides:
$\frac{v_1}{v_2} = \frac{1}{2}$
Thus,the ratio of the maximum velocities is $1 : 2$.

(C) According to Einstein's photoelectric equation,the stopping potential $V$ is given by: $eV = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
Rearranging for $V$,we get: $V = \frac{hc}{e\lambda} - \frac{\phi}{e}$.
Given that $V_1, V_2, V_3$ are in Arithmetic Progression ($A$.$P$.),we have $2V_2 = V_1 + V_3$.
Substituting the expression for $V$:
$2\left(\frac{hc}{e\lambda_2} - \frac{\phi}{e}\right) = \left(\frac{hc}{e\lambda_1} - \frac{\phi}{e}\right) + \left(\frac{hc}{e\lambda_3} - \frac{\phi}{e}\right)$.
Simplifying the equation:
$\frac{2hc}{e\lambda_2} - \frac{2\phi}{e} = \frac{hc}{e\lambda_1} + \frac{hc}{e\lambda_3} - \frac{2\phi}{e}$.
Canceling the common terms:
$\frac{2}{\lambda_2} = \frac{1}{\lambda_1} + \frac{1}{\lambda_3}$.
This condition implies that $\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \frac{1}{\lambda_3}$ are in $A$.$P$.,which means $\lambda_1, \lambda_2, \lambda_3$ are in Harmonic Progression ($H$.$P$.).

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = h\nu - \phi_0$,where $h\nu$ is the incident photon energy and $\phi_0$ is the work function.
For the first case: $2\, eV = 5\, eV - \phi_0$,which gives the work function $\phi_0 = 3\, eV$.
For the second case: The incident energy is $6\, eV$. The maximum kinetic energy of the emitted photoelectrons is $K_{\max}' = 6\, eV - 3\, eV = 3\, eV$.
To stop these photoelectrons from reaching the anode $A$,the potential of $A$ relative to $C$ $(V_{AC})$ must be negative and equal in magnitude to the maximum kinetic energy in electron volts. Thus,$V_{AC} = -3\, V$.

(B) Given: Work function $\phi = 4.5\,eV$,Wavelength $\lambda = 200\,nm$.
Energy of incident photon $E = \frac{hc}{\lambda} = \frac{1240\,eV\cdot nm}{200\,nm} = 6.2\,eV$.
Maximum kinetic energy of emitted photoelectrons $K_{max} = E - \phi = 6.2\,eV - 4.5\,eV = 1.7\,eV$.
The collector plate is at a potential of $+2.0\,V$ relative to the emitter,meaning it accelerates the electrons.
The minimum kinetic energy of electrons reaching the collector is $K_{min} = K_{initial} + qV = 0 + 2.0\,eV = 2.0\,eV$.
The maximum kinetic energy of electrons reaching the collector is $K_{max}' = K_{max} + qV = 1.7\,eV + 2.0\,eV = 3.7\,eV$.
Thus,option $B$ is correct.

(A) According to Einstein's photoelectric equation,the stopping potential $V$ is given by $eV = hf - \Phi$,where $h$ is Planck's constant,$f$ is the frequency of incident light,and $\Phi$ is the work function of the metal.
This can be rewritten as $V = (h/e)f - (\Phi/e)$.
This is the equation of a straight line $y = mx + c$,where the slope $m = h/e$ is constant for all metals.
The y-intercept is $c = -\Phi/e$.
Since $Y$ has a greater work function than $X$ $(\Phi_Y > \Phi_X)$,the magnitude of the negative y-intercept for $Y$ will be greater than that for $X$.
This means the threshold frequency $f_0 = \Phi/h$ for $Y$ will be greater than that for $X$ $(f_{0,Y} > f_{0,X})$.
Therefore,the graph for $Y$ will be shifted to the right of the graph for $X$ on the frequency axis,while both lines remain parallel.
This corresponds to the graph where $X$ is on the left and $Y$ is on the right,as shown in option $A$.

(D) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = h\nu - \Phi$, where $h$ is Planck's constant, $\nu$ is the frequency of incident light, and $\Phi$ is the work function of the emitter material.
Since the stopping potential $V_s$ is related to the maximum kinetic energy by $K_{max} = eV_s$, we have $eV_s = h\nu - \Phi$, which implies $V_s = \frac{h\nu}{e} - \frac{\Phi}{e}$.
From this equation, it is clear that the stopping potential $V_s$ depends on the frequency of the incident light $(\nu)$ and the work function $(\Phi)$, which is a property of the emitter material.
It does not depend on the intensity of the incident light.
Therefore, the correct answer is $(A)$ and $(C)$ both.

(C) The work function $\phi$ is given as $6.2 \ eV$.
The stopping potential $V_0$ is $5 \ V$,so the maximum kinetic energy $K_{max} = eV_0 = 5 \ eV$.
Using Einstein's photoelectric equation: $E = \phi + K_{max}$.
$E = 6.2 \ eV + 5 \ eV = 11.2 \ eV$.
The energy of the incident photon is $E = \frac{hc}{\lambda}$.
Using $hc \approx 1240 \ eV \cdot nm$,we get $\lambda = \frac{1240 \ eV \cdot nm}{11.2 \ eV} \approx 110.7 \ nm$.
Since the wavelength range of the ultra-violet region is approximately $10 \ nm$ to $400 \ nm$,the incident radiation lies in the ultra-violet region.

(A) Given: Wavelength $\lambda = 400 \ nm$,Energy constant $hc = 1240 \ eV \ nm$,Kinetic energy $K.E. = 1.68 \ eV$.
According to Einstein's photoelectric equation: $E = W + K.E._{max}$,where $E = \frac{hc}{\lambda}$.
Substituting the values: $E = \frac{1240}{400} = 3.1 \ eV$.
Now,calculate the work function $W$: $W = E - K.E._{max} = 3.1 \ eV - 1.68 \ eV = 1.42 \ eV$.
Note: The closest option provided is $1.41 \ eV$ (Option $A$). Assuming a slight rounding in the question's provided options,$1.42 \ eV$ is the calculated result.

(A) According to Einstein's photoelectric equation, $eV_0 = K_{max} = h\nu - \phi$, where $\phi$ is the work function of the metal surface.
Since the frequency of $X$-rays $(\nu_X)$ is much higher than the frequency of ultraviolet light $(\nu_{UV})$, the maximum kinetic energy $K_{max} = h\nu - \phi$ increases when $X$-rays are used.
Consequently, the stopping potential $V_0 = K_{max}/e$ also increases. Thus, Statement $-1$ is true.
Statement $-2$ is false because photoelectrons are emitted with a range of speeds from zero to a maximum value due to the loss of energy during collisions of electrons within the metal before they are emitted, not because of a range of frequencies in the incident light (which is monochromatic).

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ and stopping potential $V_0$ are given by: $K_{max} = eV_0 = hv - hv_0$.
Here,$h$ is Planck's constant,$v$ is the frequency of incident light,and $v_0$ is the threshold frequency.
From the equation,$K_{max} = hv - hv_0$. If the frequency $v$ is doubled to $2v$,the new maximum kinetic energy $K'_{max} = h(2v) - hv_0 = 2hv - hv_0$.
Since $2hv - hv_0$ is not equal to $2(hv - hv_0)$,the maximum kinetic energy does not double. Consequently,the stopping potential $V_0$ also does not double. Thus,Statement-$1$ is false.
Statement-$2$ is true because $K_{max}$ and $V_0$ are linear functions of frequency $v$ (i.e.,$y = mx + c$ form).

(A) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2$,where $\phi$ is the work function.
For wavelength $\lambda$: $\frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2$ --- $(1)$
For wavelength $\lambda' = \frac{3\lambda}{4}$: $\frac{hc}{\lambda'} = \frac{4hc}{3\lambda} = \phi + \frac{1}{2}mv'^2$ --- $(2)$
From $(1)$,$\frac{hc}{\lambda} = \phi + K$,where $K = \frac{1}{2}mv^2$. Thus,$\frac{hc}{3\lambda} = \frac{\phi + K}{3}$.
Substituting into $(2)$: $\frac{4}{3}(\phi + K) = \phi + \frac{1}{2}mv'^2$.
$\frac{1}{2}mv'^2 = \frac{4}{3}\phi + \frac{4}{3}K - \phi = \frac{1}{3}\phi + \frac{4}{3}K$.
Since $\phi > 0$,$\frac{1}{2}mv'^2 > \frac{4}{3}K$.
$\frac{1}{2}mv'^2 > \frac{4}{3} \left( \frac{1}{2}mv^2 \right) \implies v'^2 > \frac{4}{3}v^2$.
Therefore,$v' > v \left( \frac{4}{3} \right)^{1/2}$.

(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $K_{max} = eV_s$ ($V_s$ is the stopping potential).
For the first case: $\frac{hc}{\lambda} = 5eV_0 + \phi$ --- $(1)$
For the second case: $\frac{hc}{3\lambda} = eV_0 + \phi$ --- $(2)$
Multiply equation $(2)$ by $5$: $\frac{5hc}{3\lambda} = 5eV_0 + 5\phi$ --- $(3)$
Subtract equation $(1)$ from equation $(3)$:
$\frac{5hc}{3\lambda} - \frac{hc}{\lambda} = (5eV_0 + 5\phi) - (5eV_0 + \phi)$
$\frac{5hc - 3hc}{3\lambda} = 4\phi$
$\frac{2hc}{3\lambda} = 4\phi$
$\phi = \frac{2hc}{12\lambda} = \frac{hc}{6\lambda}$

(B) The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{1242 \ eV \ nm}{400 \ nm} = 3.105 \ eV$.
$(a)$ For the photoelectric effect to occur,the energy of the incident photon must be greater than the work function $(\phi)$. Here,$E = 3.105 \ eV$,which is less than the work functions of $Zn$ $(3.4 \ eV)$,$Fe$ $(4.8 \ eV)$,and $Ni$ $(5.9 \ eV)$. Therefore,no photoelectrons will be emitted from any metal. Statement $(a)$ is incorrect.
$(b)$ Since $E < \phi$ for all metals,no photoelectrons will be emitted from $Ni$. Statement $(b)$ is correct.
$(c)$ According to Einstein's photoelectric equation,$KE_{max} = E - \phi = h\nu - \phi$. If the frequency $\nu$ is doubled,$KE_{max}' = 2h\nu - \phi$. This is not equal to $2(h\nu - \phi)$. Statement $(c)$ is incorrect.
$(d)$ If $\lambda < 200 \ nm$,then $E = \frac{1242}{\lambda} > \frac{1242}{200} = 6.21 \ eV$. Since $6.21 \ eV$ is greater than the work functions of all three metals $(3.4 \ eV, 4.8 \ eV, 5.9 \ eV)$,photoelectrons will be emitted from all three metals. Statement $(d)$ is correct.
Thus,the correct statements are $(b)$ and $(d)$.

(A) 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 surface.
The energy of the incident photon $E$ is calculated as $E = \frac{12400}{\lambda(\text{in } \mathring{A})} \ eV$.
Given $\lambda = 2000\ \mathring{A}$,we have $E = \frac{12400}{2000} = 6.2\ eV$.
The work function $\phi = 5.01\ eV$.
Therefore,$K_{max} = 6.2\ eV - 5.01\ eV = 1.19\ eV \approx 1.2\ eV$.
The stopping potential $V_S$ is related to the maximum kinetic energy by $K_{max} = eV_S$.
Thus,$eV_S = 1.2\ eV$,which implies $V_S = 1.2\ V$.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $hc \approx 1240 \ eV \cdot nm$,we have $E = \frac{1240 \ eV \cdot nm}{200 \ nm} = 6.2 \ eV$.
The maximum kinetic energy $(K_{\max})$ of the emitted photoelectrons is given by Einstein's photoelectric equation: $K_{\max} = E - \phi$,where $\phi$ is the work function.
$K_{\max} = 6.2 \ eV - 4.7 \ eV = 1.5 \ eV$.
The maximum potential $(V_{\max})$ acquired by the ball is equal to the stopping potential,which is $V_{\max} = \frac{K_{\max}}{e} = 1.5 \ V$.

(D) According to Einstein's photoelectric equation: $eV = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function.
Case $I$: $eV = \frac{hc}{\lambda} - \phi$ --- $(i)$
Case $II$: $e(V/3) = \frac{hc}{2\lambda} - \phi$ --- $(ii)$
Multiply equation $(ii)$ by $3$: $eV = \frac{3hc}{2\lambda} - 3\phi$ --- $(iii)$
Equating $(i)$ and $(iii)$: $\frac{hc}{\lambda} - \phi = \frac{3hc}{2\lambda} - 3\phi$
$2\phi = \frac{3hc}{2\lambda} - \frac{hc}{\lambda} = \frac{hc}{2\lambda}$
$\phi = \frac{hc}{4\lambda}$
Since $\phi = \frac{hc}{\lambda_0}$,we get $\lambda_0 = 4\lambda$.

(B) The photoelectric equation is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$.
Given: $\lambda_0 = 400 \ nm = 4000 \ \mathring{A}$,$K_{max} = 0.9 \ eV$.
Using the relation $E (\text{in } eV) = \frac{12400}{\lambda (\text{in } \mathring{A})}$,the work function $\phi = \frac{12400}{4000} = 3.1 \ eV$.
Substituting the values into the photoelectric equation:
$0.9 = \frac{12400}{\lambda} - 3.1$
$0.9 + 3.1 = \frac{12400}{\lambda}$
$4.0 = \frac{12400}{\lambda}$
$\lambda = \frac{12400}{4} = 3100 \ \mathring{A} = 310 \ nm$.

(D) According to Einstein's photoelectric equation, $K_{max} = h\nu - \phi$, where $K_{max}$ is the maximum kinetic energy of emitted photoelectrons, $h$ is Planck's constant, $\nu$ is the frequency of incident radiation, and $\phi$ is the work function of the metal.
Increasing the frequency $(\nu)$ of the incident radiation increases the maximum kinetic energy $(K_{max})$ of the emitted photoelectrons. Consequently, the stopping potential $(V_s)$, which is defined as $V_s = K_{max}/e$, becomes more negative (shifts to the left on the potential axis).
Intensity of incident radiation is directly proportional to the number of photons striking the surface per unit time, which in turn determines the saturation photocurrent. Reducing the intensity decreases the number of photoelectrons emitted per unit time, thereby reducing the saturation photocurrent.
Comparing the given curves with the broken line curve: Curve $D$ shows a more negative stopping potential (due to increased frequency) and a lower saturation current (due to decreased intensity). Therefore, curve $D$ represents the new situation.

(B) The stopping potential $(V_s)$ depends only on the frequency (or wavelength) of the incident light and the work function of the metal surface,as given by Einstein's photoelectric equation: $eV_s = h\nu - \phi$.
It does not depend on the intensity of the incident light.
When the distance of the source is increased from $10\, \text{cm}$ to $20\, \text{cm}$,the intensity of the light incident on the surface decreases due to the inverse square law.
However,the wavelength $(\lambda)$ and the frequency $(\nu)$ of the incident photons remain unchanged.
Since the stopping potential is independent of the intensity,it will remain the same.
Therefore,the stopping potential remains $1.5\, \text{V}$.

(B) The energy of a photon is given by $E = \frac{hc}{\lambda}$. Using $hc \approx 1240 \ eV \cdot nm$:
For $\lambda_1 = 310 \ nm$,$E_1 = \frac{1240}{310} = 4.0 \ eV$.
For $\lambda_2 = 455 \ nm$,$E_2 = \frac{1240}{455} \approx 2.72 \ eV$.
For $\lambda_3 = 620 \ nm$,$E_3 = \frac{1240}{620} = 2.0 \ eV$.
According to Einstein's photoelectric equation,$K_{max} = E - \Phi$,where $\Phi = 1.2 \ eV$.
For $\lambda_1 = 310 \ nm$,$K_{max} = 4.0 - 1.2 = 2.8 \ eV$.
For $\lambda_2 = 455 \ nm$,$K_{max} = 2.72 - 1.2 = 1.52 \ eV$.
For $\lambda_3 = 620 \ nm$,$K_{max} = 2.0 - 1.2 = 0.8 \ eV$.
The maximum kinetic energy is the highest value among these,which is $2.8 \ eV$.

(C) From the given $I - V$ curve,the stopping potential $V_s = 5 \, V$ and the saturation current $I_s = 10 \, \mu A = 10^{-5} \, A$.
The energy of incident photons is given by Einstein's photoelectric equation:
$E = h\nu = eV_s + \phi_0$
$E = 5 \, eV + 2 \, eV = 7 \, eV$.
The saturation current is related to the power $P$ of the incident light by:
$I_s = \eta \times \left( \frac{P}{E} \right) \times e$
where $\eta$ is the efficiency of photoemission $= 10^{-3} \% = 10^{-5}$.
Substituting the values:
$10^{-5} = 10^{-5} \times \left( \frac{P}{7 \, eV} \right) \times e$
$1 = \frac{P}{7 \, eV} \times e$
$1 = \frac{P}{7}$
$P = 7 \, W$.

(B) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi = eV$.
For the first case: $\frac{hc}{\lambda} = \phi + eV$ .......$(1)$
For the second case: $\frac{hc}{2\lambda} = \phi + \frac{eV}{3}$ .......$(2)$
Multiply equation $(2)$ by $3$: $\frac{3hc}{2\lambda} = 3\phi + eV$ .......$(3)$
Subtract equation $(1)$ from equation $(3)$: $(\frac{3}{2} - 1) \frac{hc}{\lambda} = (3\phi - \phi) + (eV - eV)$.
$\frac{1}{2} \frac{hc}{\lambda} = 2\phi$.
$\phi = \frac{hc}{4\lambda}$.
Since $\phi = \frac{hc}{\lambda_{th}}$,we get $\lambda_{th} = 4\lambda$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of an electron emitted from plate $1$ is $K_{max} = hv - \phi$.
When the electron moves from plate $1$ to plate $2$,it is accelerated by the potential difference between the plates.
The battery is connected such that plate $1$ is at a lower potential than plate $2$ (or the circuit configuration provides an accelerating potential). The potential difference $V$ across the capacitor is equal to the $emf$ of the battery,which is $V = \frac{\phi}{e}$.
The work done by the electric field on the electron is $W = eV = e \cdot \frac{\phi}{e} = \phi$.
The maximum kinetic energy of the electron reaching plate $2$ is the sum of its initial maximum kinetic energy and the work done by the electric field:
$K_{final} = K_{max} + W = (hv - \phi) + \phi = hv$.
Therefore,the correct option is $D$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by:
$K_{max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$
When a charged particle moves in a magnetic field $B_0$ in a circular path of radius $r$,the magnetic force provides the centripetal force:
$evB_0 = \frac{mv^2}{r} \implies v = \frac{eB_0r}{m}$
The kinetic energy can be expressed as:
$K_{max} = \frac{1}{2}mv^2 = \frac{1}{2}m \left( \frac{eB_0r}{m} \right)^2 = \frac{e^2 B_0^2 r^2}{2m}$
Equating the two expressions for $K_{max}$:
$\frac{e^2 B_0^2 r^2}{2m} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$
Rearranging to solve for $\frac{1}{\lambda_0}$:
$\frac{hc}{\lambda_0} = \frac{hc}{\lambda} - \frac{e^2 B_0^2 r^2}{2m}$
Dividing both sides by $hc$:
$\frac{1}{\lambda_0} = \frac{1}{\lambda} - \frac{e^2 B_0^2 r^2}{2mhc}$

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE_{\max})$ of emitted photo-electrons is given by:
$KE_{\max} = E - \phi$
where $E$ is the energy of the incident photon and $\phi$ is the work function of the metal.
Given: $E = 6\, eV$ and $\phi = 4\, eV$.
$KE_{\max} = 6\, eV - 4\, eV = 2\, eV$.
The kinetic energy of emitted photo-electrons ranges from $0$ to $KE_{\max}$.
Therefore,the minimum kinetic energy $(KE_{\min})$ of the emitted photo-electrons is $0\, eV$.

(C) According to Einstein's photoelectric equation,the kinetic energy $K$ is given by: $K = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
For the initial state: $K = \frac{hc}{\lambda} - \phi$ --- $(1)$
For the final state,where kinetic energy is $2K$: $2K = \frac{hc}{\lambda'} - \phi$ --- $(2)$
From $(1)$,we have $\frac{hc}{\lambda} = K + \phi$. Since $K > 0$,it follows that $\frac{hc}{\lambda} > \phi$.
From $(2)$,$\frac{hc}{\lambda'} = 2K + \phi$. Since $K > 0$,it follows that $\frac{hc}{\lambda'} > \phi$.
Dividing $(1)$ by $(2)$: $\frac{\lambda'}{\lambda} = \frac{K + \phi}{2K + \phi}$.
Since $K + \phi < 2K + \phi$,the ratio $\frac{\lambda'}{\lambda} < 1$,which implies $\lambda' < \lambda$.
Also,$\frac{\lambda'}{\lambda} = \frac{K + \phi}{2K + \phi} = \frac{1}{2} \left( \frac{2K + 2\phi}{2K + \phi} \right) = \frac{1}{2} \left( \frac{2K + \phi + \phi}{2K + \phi} \right) = \frac{1}{2} \left( 1 + \frac{\phi}{2K + \phi} \right)$.
Since $\frac{\phi}{2K + \phi} > 0$,the term in the parenthesis is greater than $1$. Therefore,$\frac{\lambda'}{\lambda} > \frac{1}{2}$,which implies $\lambda' > \lambda/2$.
Combining these,we get $\lambda > \lambda' > \lambda/2$.

(C) The given electric field equation is $\varepsilon = (100 \text{ V/m}) [\sin(\omega_1 t) + \sin(\omega_2 t)]$, where $\omega_1 = 5 \times 10^{15} \text{ rad/s}$ and $\omega_2 = 8 \times 10^{15} \text{ rad/s}$.
To find the maximum kinetic energy, we must consider the photon with the highest frequency, as $K.E._{\max} = h\nu - \phi$.
The frequency $\nu$ is given by $\nu = \frac{\omega}{2\pi}$. For the higher frequency component, $\nu = \frac{8 \times 10^{15}}{2\pi} \text{ Hz}$.
The energy of the photon $E = h\nu = \frac{6.626 \times 10^{-34} \times 8 \times 10^{15}}{2 \times 3.1416} \text{ Joules}$.
Converting this to $\text{eV}$ by dividing by $1.6 \times 10^{-19} \text{ C}$:
$E = \frac{6.626 \times 10^{-34} \times 8 \times 10^{15}}{2 \times 3.1416 \times 1.6 \times 10^{-19}} \approx 5.27 \text{ eV}$.
The maximum kinetic energy is $K.E._{\max} = E - \phi = 5.27 \text{ eV} - 2 \text{ eV} = 3.27 \text{ eV}$.

(B) According to Einstein's photoelectric equation:
$hv_1 = hv_0 + \frac{1}{2}mv_1^2$ --- $(1)$
$hv_2 = hv_0 + \frac{1}{2}mv_2^2$ --- $(2)$
Subtracting equation $(2)$ from $(1)$:
$h(v_1 - v_2) = \frac{1}{2}m(v_1^2 - v_2^2)$
Rearranging the terms to solve for the velocity difference:
$v_1^2 - v_2^2 = \frac{2h}{m}(v_1 - v_2)$
Thus,option $B$ is correct.

(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 material.
Given: $E = 25\, eV$ and $\Phi = 7\, eV$.
$K_{max} = 25\, eV - 7\, eV = 18\, eV$.
The stopping potential $(V_0)$ is related to the maximum kinetic energy by the equation $K_{max} = eV_0$.
Therefore,$eV_0 = 18\, eV$,which implies $V_0 = 18\, V$.

(D) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + \frac{1}{2}mv_{\max}^2$.
Let the initial wavelength be $\lambda_1 = \lambda$ and the final wavelength be $\lambda_2 = 4\lambda$.
The maximum kinetic energy is given by $K_{\max} = \frac{hc}{\lambda} - W_0$.
Since $K_{\max} = \frac{1}{2}mv_{\max}^2$,we have $v_{\max} = \sqrt{\frac{2}{m} (\frac{hc}{\lambda} - W_0)}$.
When $\lambda$ increases to $4\lambda$,the energy of the incident photon $\frac{hc}{4\lambda}$ decreases.
Since $K_{\max}$ decreases,$v_{\max}$ also decreases.
Let the new velocity be $v'$. Then $v' = \sqrt{\frac{2}{m} (\frac{hc}{4\lambda} - W_0)}$.
Comparing $v'$ with $v_{\max}$: $v' = \sqrt{\frac{1}{4} \frac{2hc}{\lambda} - \frac{2W_0}{m}} = \sqrt{\frac{1}{4} v_{\max}^2 - \frac{3W_0}{2m}}$.
Since $\frac{1}{4} v_{\max}^2 - \frac{3W_0}{2m} < \frac{1}{4} v_{\max}^2$,it follows that $v' < \frac{1}{2} v_{\max}$.
Thus,the factor by which the velocity changes is less than $\frac{1}{2}$.

(A) According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by $K_{max} = eV_0$, where $V_0$ is the stopping potential.
Einstein's equation is $K_{max} = \frac{hc}{\lambda} - \phi$.
Substituting $K_{max} = eV_0$, we get $eV_0 = \frac{hc}{\lambda} - \phi$.
Given: Work function $\phi = 2.14 \ eV$, Stopping potential $V_0 = 0.60 \ V$.
Using the relation $\frac{hc}{\lambda} \approx \frac{12400 \ eV \cdot \mathring{A}}{\lambda}$, we have $0.60 \ eV = \frac{12400 \ eV \cdot \mathring{A}}{\lambda} - 2.14 \ eV$.
$0.60 + 2.14 = \frac{12400}{\lambda}$.
$2.74 = \frac{12400}{\lambda}$.
$\lambda = \frac{12400}{2.74} \approx 4525.5 \ \mathring{A}$.
Converting to nanometers, $\lambda \approx 452.55 \ nm$, which is approximately $454 \ nm$ when using $hc = 1242 \ eV \cdot nm$ or similar constants.

(B) From the given graph, the stopping potential $V_{0}$ is the negative voltage at which the photocurrent becomes zero.
Looking at the graph, the curve intersects the voltage axis at $V = -4\,V$.
Therefore, the magnitude of the stopping potential is $|V_{0}| = 4\,V$.
The maximum kinetic energy $K_{\max}$ of the emitted photoelectrons is given by the formula $K_{\max} = e|V_{0}|$.
Substituting the value of $|V_{0}|$, we get $K_{\max} = e \times 4\,V = 4\,eV$.

(C) The power of the light source is $P = 5 \times 10^{-3} \, W$ and the number of photons emitted per second is $n = 8 \times 10^{15} \, s^{-1}$.
The energy of a single photon $E$ is given by $E = \frac{P}{n}$.
$E = \frac{5 \times 10^{-3}}{8 \times 10^{15}} \, J = 6.25 \times 10^{-19} \, J$.
To convert this energy into electron-volts $(eV)$,divide by the charge of an electron $e = 1.6 \times 10^{-19} \, C$:
$E = \frac{6.25 \times 10^{-19}}{1.6 \times 10^{-19}} \, eV = 3.90625 \, eV \approx 3.9 \, eV$.
According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E)_{\max}$ is related to the stopping potential $V_s$ as $(K.E)_{\max} = e V_s = 2.0 \, eV$.
The photoelectric equation is $E = \phi + (K.E)_{\max}$,where $\phi$ is the work function.
$\phi = E - (K.E)_{\max} = 3.9 \, eV - 2.0 \, eV = 1.9 \, eV$.
Therefore,the work function of the metal is $1.9 \, eV$.