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

(C) According to Einstein's photoelectric equation,the kinetic energy $E$ of the ejected electrons is given by: $E = \frac{hc}{\lambda} - \Phi$,where $\Phi$ is the work function of the surface.
For the initial state: $E = \frac{hc}{\lambda} - \Phi$.
For the final state,the energy becomes $2E$: $2E = \frac{hc}{\lambda_1} - \Phi$.
From the first equation,$\Phi = \frac{hc}{\lambda} - E$.
Substituting this into the second equation: $2E = \frac{hc}{\lambda_1} - (\frac{hc}{\lambda} - E)$.
$2E = \frac{hc}{\lambda_1} - \frac{hc}{\lambda} + E$.
$E = \frac{hc}{\lambda_1} - \frac{hc}{\lambda}$.
Since $E > 0$,it follows that $\frac{hc}{\lambda_1} > \frac{hc}{\lambda}$,which implies $\lambda_1 < \lambda$.
Also,$\frac{hc}{\lambda_1} = E + \frac{hc}{\lambda}$.
Since $E = \frac{hc}{\lambda} - \Phi$,we have $\frac{hc}{\lambda_1} = \frac{hc}{\lambda} - \Phi + \frac{hc}{\lambda} = \frac{2hc}{\lambda} - \Phi$.
Since $\Phi > 0$,$\frac{hc}{\lambda_1} < \frac{2hc}{\lambda}$,which implies $\lambda_1 > \frac{\lambda}{2}$.
Thus,$\frac{\lambda}{2} < \lambda_1 < \lambda$.

(B) The kinetic energy of the emitted photoelectrons is given by $K_{max} = \frac{1}{2}mv^2$.
At the stopping potential $(V_s)$,the work done by the retarding potential equals the maximum kinetic energy: $eV_s = \frac{1}{2}mv^2$.
Rearranging for the stopping potential,we get $V_s = \frac{1}{2} \left(\frac{m}{e}\right) v^2$.
Given $\frac{e}{m} = 1.8 \times 10^{11} \ C/kg$,therefore $\frac{m}{e} = \frac{1}{1.8 \times 10^{11}} \ kg/C$.
Substituting the values: $V_s = \frac{1}{2} \times \frac{1}{1.8 \times 10^{11}} \times (9 \times 10^5)^2$.
$V_s = \frac{1}{2} \times \frac{81 \times 10^{10}}{1.8 \times 10^{11}} = \frac{81}{3.6} = 22.5 \times 0.1 = 2.25 \ V$.
Thus,the stopping potential is $2.25 \ V$.

(C) The energy of an incident photon is given by $E = \frac{hc}{\lambda}$.
For the photoelectric effect to occur,the energy of the incident photon must be greater than or equal to the work function $(\Phi)$ of the metal.
The work function is given as $\Phi = \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.
Therefore,the condition for the photoelectric effect is $E \geqslant \Phi$.
Substituting the expressions,we get $\frac{hc}{\lambda} \geqslant \frac{hc}{\lambda_0}$.
Dividing both sides by $hc$,we get $\frac{1}{\lambda} \geqslant \frac{1}{\lambda_0}$.
Taking the reciprocal,the inequality sign reverses: $\lambda \leqslant \lambda_0$.
Since the standard condition for emission is that the incident wavelength must be less than or equal to the threshold wavelength,the correct option is $\lambda < \lambda_0$ (or $\lambda \leqslant \lambda_0$).

(B) According to Einstein's photoelectric equation: $K_{max} = h\nu - \Phi$,where $\Phi$ is the work function.
Initially: $0.4 = h\nu - \Phi$ --- $(1)$
When frequency is increased by $20 \%$,the new frequency $\nu' = 1.2\nu$.
The new kinetic energy is: $0.7 = h(1.2\nu) - \Phi$ --- $(2)$
From $(1)$,$h\nu = 0.4 + \Phi$.
Substitute this into $(2)$: $0.7 = 1.2(0.4 + \Phi) - \Phi$.
$0.7 = 0.48 + 1.2\Phi - \Phi$.
$0.7 - 0.48 = 0.2\Phi$.
$0.22 = 0.2\Phi$.
$\Phi = \frac{0.22}{0.2} = 1.1 \ eV$.

(D) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = h v - \Phi_0$, where $h$ is Planck's constant, $v$ is the frequency of incident light, and $\Phi_0 = h v_0$ is the work function ($v_0$ is the threshold frequency).
For frequency $v_1$, $K_1 = h v_1 - h v_0 = h(v_1 - v_0)$.
For frequency $v_2$, $K_2 = h v_2 - h v_0 = h(v_2 - v_0)$.
Given the ratio $K_1 : K_2 = 1 : k$, we have $\frac{K_1}{K_2} = \frac{1}{k}$.
Substituting the expressions: $\frac{h(v_1 - v_0)}{h(v_2 - v_0)} = \frac{1}{k}$.
This simplifies to $k(v_1 - v_0) = v_2 - v_0$.
Expanding the terms: $k v_1 - k v_0 = v_2 - v_0$.
Rearranging to solve for $v_0$: $k v_1 - v_2 = k v_0 - v_0 = v_0(k - 1)$.
Therefore, $v_0 = \frac{k v_1 - v_2}{k - 1}$.

(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $K_{max} = \frac{1}{2}mv^2$.
For wavelength $\lambda$: $\frac{1}{2}mV^2 = \frac{hc}{\lambda} - \phi$ --- $(1)$
For wavelength $\lambda' = \frac{2\lambda}{3}$: $\frac{1}{2}mv'^2 = \frac{hc}{2\lambda/3} - \phi = \frac{3hc}{2\lambda} - \phi$ --- $(2)$
From $(1)$,$\frac{hc}{\lambda} = \frac{1}{2}mV^2 + \phi$.
Substitute this into $(2)$: $\frac{1}{2}mv'^2 = \frac{3}{2}(\frac{1}{2}mV^2 + \phi) - \phi = \frac{3}{4}mV^2 + \frac{1}{2}\phi$.
Since $\phi > 0$,$\frac{1}{2}mv'^2 > \frac{3}{4}mV^2$.
$v'^2 > \frac{3}{2}V^2$,which implies $v' > \sqrt{\frac{3}{2}}V$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of 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.
Case $1$: When incident energy $E_1 = 2\Phi$,the maximum kinetic energy is $K_1 = 2\Phi - \Phi = \Phi$.
Since $K_1 = \frac{1}{2} m V_1^2$,we have $\frac{1}{2} m V_1^2 = \Phi$,which implies $V_1 = \sqrt{\frac{2\Phi}{m}}$.
Case $2$: When incident energy $E_2 = 3\Phi$,the maximum kinetic energy is $K_2 = 3\Phi - \Phi = 2\Phi$.
Since $K_2 = \frac{1}{2} m V_2^2$,we have $\frac{1}{2} m V_2^2 = 2\Phi$,which implies $V_2 = \sqrt{\frac{4\Phi}{m}}$.
Taking the ratio $V_1: V_2$:
$\frac{V_1}{V_2} = \frac{\sqrt{\frac{2\Phi}{m}}}{\sqrt{\frac{4\Phi}{m}}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Therefore,the ratio $V_1: V_2$ is $1: \sqrt{2}$.

(C) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by: $eV_s = \frac{hc}{\lambda} - \Phi$,where $\Phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength.
For the first case: $eV_1 = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_0} \right)$ --- $(1)$
For the second case: $e \left( \frac{V_1}{6} \right) = \frac{hc}{3\lambda_1} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{3\lambda_1} - \frac{1}{\lambda_0} \right)$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$6 = \frac{\frac{1}{\lambda_1} - \frac{1}{\lambda_0}}{\frac{1}{3\lambda_1} - \frac{1}{\lambda_0}}$
$6 \left( \frac{1}{3\lambda_1} - \frac{1}{\lambda_0} \right) = \frac{1}{\lambda_1} - \frac{1}{\lambda_0}$
$\frac{2}{\lambda_1} - \frac{6}{\lambda_0} = \frac{1}{\lambda_1} - \frac{1}{\lambda_0}$
$\frac{2}{\lambda_1} - \frac{1}{\lambda_1} = \frac{6}{\lambda_0} - \frac{1}{\lambda_0}$
$\frac{1}{\lambda_1} = \frac{5}{\lambda_0}$
$\lambda_0 = 5\lambda_1$. Thus,the correct option is $C$.

(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.
Given the threshold frequency is $v$,the work function is $\Phi = hv$.
The energy of the incident radiation with frequency $4v$ is $E = h(4v) = 4hv$.
Substituting these values into the equation:
$K_{max} = 4hv - hv = 3hv$.
Since $K_{max} = \frac{1}{2}mv_{max}^2$,we have:
$\frac{1}{2}mv_{max}^2 = 3hv$.
Solving for $v_{max}$:
$v_{max}^2 = \frac{6hv}{m}$.
$v_{max} = \sqrt{\frac{6hv}{m}}$.

(A) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by:
$eV_s = h\nu - \phi$
$V_s = (h/e)\nu - (\phi/e)$
Comparing this with the equation of a straight line $y = mx + c$,we get the slope $m = h/e$.
Here,$h$ is Planck's constant and $e$ is the charge of an electron.
Since both $h$ and $e$ are universal constants,the slope of the graph of stopping potential versus frequency is independent of the work function $\phi$ of the material.
Therefore,the slope for both materials is the same,i.e.,$h/e$.
Thus,the ratio of the slopes is $1: 1$.

(A) According to Einstein's photoelectric equation,the stopping potential $V_s$ is related to the frequency $\nu$ of incident radiation by the equation:
$eV_s = h\nu - \phi$
where $h$ is Planck's constant,$e$ is the charge of an electron,and $\phi$ is the work function of the metal.
Rearranging for $V_s$,we get:
$V_s = \frac{h}{e}\nu - \frac{\phi}{e}$
This is the equation of a straight line $y = mx + c$,where the slope $m = \frac{h}{e}$.
From the given graph,the slope of the line can be calculated using two points $( \nu_1, V_1 )$ and $( \nu_2, V_2 )$:
Slope $= \frac{V_2 - V_1}{\nu_2 - \nu_1}$
Equating the two expressions for the slope:
$\frac{h}{e} = \frac{V_2 - V_1}{\nu_2 - \nu_1}$
Therefore,the value of Planck's constant $h$ is:
$h = \frac{e(V_2 - V_1)}{\nu_2 - \nu_1}$

(D) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \Phi + K_{max}$,where $\Phi$ is the work function and $K_{max} = \frac{1}{2}mv^2$.
For wavelength $\lambda_1$ with velocity $V$: $\frac{hc}{\lambda_1} = \Phi + \frac{1}{2}mV^2$ --- $(1)$
For wavelength $\lambda_2$ with velocity $2V$: $\frac{hc}{\lambda_2} = \Phi + \frac{1}{2}m(2V)^2 = \Phi + 2mV^2$ --- $(2)$
From $(1)$,$\frac{1}{2}mV^2 = \frac{hc}{\lambda_1} - \Phi$. Multiplying by $4$,we get $2mV^2 = \frac{4hc}{\lambda_1} - 4\Phi$.
Substitute this into $(2)$: $\frac{hc}{\lambda_2} = \Phi + \frac{4hc}{\lambda_1} - 4\Phi$.
$\frac{hc}{\lambda_2} = \frac{4hc}{\lambda_1} - 3\Phi$.
$3\Phi = \frac{4hc}{\lambda_1} - \frac{hc}{\lambda_2} = hc \left( \frac{4\lambda_2 - \lambda_1}{\lambda_1 \lambda_2} \right)$.
$\Phi = \frac{hc}{3 \lambda_1 \lambda_2} (4 \lambda_2 - \lambda_1)$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the 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 metal.
Since the stopping potential $V_s$ is related to the maximum kinetic energy by $K_{max} = eV_s$,we have $eV_s = h\nu - \phi$,or $V_s = \frac{h\nu}{e} - \frac{\phi}{e}$.
Initially,let the frequency be $\nu_1$ and stopping potential be $V_{s1} = \frac{h\nu_1}{e} - \frac{\phi}{e}$.
When the frequency is doubled,$\nu_2 = 2\nu_1$,the new stopping potential is $V_{s2} = \frac{h(2\nu_1)}{e} - \frac{\phi}{e} = \frac{2h\nu_1}{e} - \frac{\phi}{e}$.
Comparing $V_{s2}$ with $V_{s1}$,we see that $V_{s2} = 2V_{s1} + \frac{\phi}{e}$.
Since $\frac{\phi}{e} > 0$,it follows that $V_{s2} > 2V_{s1}$.
Therefore,the stopping potential becomes more than double.

(D) The threshold frequency of the material is $\nu_0$. The initial incident frequency is $\nu_1 = 3\nu_0$.
When the incident frequency is changed to $\nu_2 = \frac{1}{4} \nu_1 = \frac{1}{4} (3\nu_0) = 0.75 \nu_0$.
For the photoelectric effect to occur,the incident frequency must be greater than or equal to the threshold frequency $(\nu \ge \nu_0)$.
Since the new frequency $\nu_2 = 0.75 \nu_0$ is less than the threshold frequency $\nu_0$,no photoelectrons will be emitted regardless of the intensity of the incident light.
Therefore,the photoelectric current will be zero.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = E - W$,where $E$ is the energy of the incident photon and $W$ is the work function.
Case $1$: $E_1 = 4W$.
$K_1 = 4W - W = 3W$.
Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv_1^2 = 3W$,so $v_1 = \sqrt{\frac{6W}{m}}$.
Case $2$: $E_2 = 6W$.
$K_2 = 6W - W = 5W$.
Similarly,$\frac{1}{2}mv_2^2 = 5W$,so $v_2 = \sqrt{\frac{10W}{m}}$.
The ratio of velocities is $\frac{v_1}{v_2} = \frac{\sqrt{6W/m}}{\sqrt{10W/m}} = \sqrt{\frac{6}{10}} = \sqrt{\frac{3}{5}} = \sqrt{3}: \sqrt{5}$.

(C) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \phi$, where $E$ is the incident photon energy and $\phi$ is the work function.
For the first case: $E_1 = 3\phi$. Thus, $K_{max1} = 3\phi - \phi = 2\phi$.
We know $K_{max} = \frac{1}{2}mv^2$, so $\frac{1}{2}mv_1^2 = 2\phi$ --- $(1)$.
For the second case: $E_2 = 7\phi$. Thus, $K_{max2} = 7\phi - \phi = 6\phi$.
So, $\frac{1}{2}mv_2^2 = 6\phi$ --- $(2)$.
Dividing equation $(2)$ by equation $(1)$: $\frac{v_2^2}{v_1^2} = \frac{6\phi}{2\phi} = 3$.
Therefore, $v_2 = v_1 \sqrt{3}$.
Given $v_1 = 4 \times 10^6 \,m/s$, we get $v_2 = 4 \sqrt{3} \times 10^6 \,m/s$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ 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.
For metal $A$: $K_{A} = hf - \phi_A$
For metal $B$: $K_{B} = h(2f) - \phi_B = 2hf - \phi_B$
Given the ratio of work functions $\frac{\phi_A}{\phi_B} = \frac{1}{2}$,we have $\phi_B = 2\phi_A$.
Substituting this into the expression for $K_B$: $K_B = 2hf - 2\phi_A = 2(hf - \phi_A)$.
Now,the ratio of kinetic energies is $\frac{K_A}{K_B} = \frac{hf - \phi_A}{2(hf - \phi_A)} = \frac{1}{2}$.

(C) According to Einstein's photoelectric equation,$K_{max} = eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.
For the first case: $\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = eV$ $(i)$
For the second case: $\frac{hc}{3\lambda} - \frac{hc}{\lambda_0} = e\left(\frac{V}{6}\right)$ (ii)
Dividing equation $(i)$ by equation (ii):
$\frac{\frac{hc}{\lambda} - \frac{hc}{\lambda_0}}{\frac{hc}{3\lambda} - \frac{hc}{\lambda_0}} = \frac{eV}{eV/6} = 6$
$\frac{1}{\lambda} - \frac{1}{\lambda_0} = 6 \left( \frac{1}{3\lambda} - \frac{1}{\lambda_0} \right)$
$\frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{2}{\lambda} - \frac{6}{\lambda_0}$
$\frac{6}{\lambda_0} - \frac{1}{\lambda_0} = \frac{2}{\lambda} - \frac{1}{\lambda}$
$\frac{5}{\lambda_0} = \frac{1}{\lambda}$
$\lambda_0 = 5\lambda$

(C) The maximum kinetic energy $(K.E.)_{\max}$ of the ejected electrons is related to the stopping potential $V_s$ by the equation:
$(K.E.)_{\max} = e V_s$
Given that the stopping potential $V_s = 10 \ V$ and the charge of an electron $e = 1.6 \times 10^{-19} \ C$,we substitute these values into the equation:
$(K.E.)_{\max} = (1.6 \times 10^{-19} \ C) \times (10 \ V)$
$(K.E.)_{\max} = 1.6 \times 10^{-18} \ J$
Therefore,the maximum kinetic energy is $1.6 \times 10^{-18} \ J$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $k$ is given by:
$k = h\nu - E_0$
From this,we can express the work function $E_0$ as:
$E_0 = h\nu - k$
When the frequency of the incident radiation is doubled $(2\nu)$,let the new maximum kinetic energy be $k'$. The new equation becomes:
$k' = h(2\nu) - E_0$
Substituting the value of $E_0$ into the equation:
$k' = 2h\nu - (h\nu - k)$
$k' = 2h\nu - h\nu + k$
$k' = h\nu + k$

(C) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K.E_{\max} = h\nu - W$,where $h\nu$ is the energy of the incident photon and $W$ is the work function of the material.
Initially,$K_1 = h\nu - W$.
When the frequency is doubled,the new frequency becomes $2\nu$. The new maximum kinetic energy $K_2$ is:
$K_2 = h(2\nu) - W = 2h\nu - W$.
We can rewrite this as:
$K_2 = 2(h\nu - W) + W = 2K_1 + W$.
Since the work function $W$ is a positive constant,$K_2 > 2K_1$.
Therefore,the kinetic energy of the emitted photoelectrons will be more than two times its initial value.

(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by:
$e V_s = h \nu - \phi$
$V_s = \frac{h}{e} \nu - \frac{\phi}{e}$
Comparing this with the equation of a straight line $y = mx + c$,the intercept on the frequency axis (where $V_s = 0$) is the threshold frequency $\nu_0$,given by $\nu_0 = \frac{\phi}{h}$,which implies $\phi = h \nu_0$.
From the graph,it is clear that the threshold frequency for metal $X$ $(\nu_0)$ is less than the threshold frequency for metal $Y$ $(\nu_0^{\prime})$,i.e.,$\nu_0 < \nu_0^{\prime}$.
Since the work function $\phi$ is directly proportional to the threshold frequency $\nu_0$,we have $\phi_x < \phi_y$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by:
$K_{max} = \frac{hc}{\lambda} - W_0$
We also know that the maximum kinetic energy is related to the stopping potential $V$ by the equation:
$K_{max} = eV$
Equating the two expressions for $K_{max}$:
$eV = \frac{hc}{\lambda} - W_0$
Rearranging the equation to solve for $\lambda$:
$\frac{hc}{\lambda} = W_0 + eV$
$\lambda = \frac{hc}{W_0 + eV}$

(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K.E._{\max} = hF - \Phi$,where $\Phi = hF_0$ is the work function.
Case $1$: When incident frequency $F = 2F_0$,
$\frac{1}{2}mV_1^2 = h(2F_0) - hF_0 = hF_0$
Case $2$: When incident frequency $F = 5F_0$,
$\frac{1}{2}mV_2^2 = h(5F_0) - hF_0 = 4hF_0$
Dividing the two equations:
$\frac{V_1^2}{V_2^2} = \frac{hF_0}{4hF_0} = \frac{1}{4}$
Taking the square root on both sides:
$\frac{V_1}{V_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

(A) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K.E_{\max} = E - \phi$.
For the first case,$E_1 = 3\phi$,so $K.E_1 = 3\phi - \phi = 2\phi$.
For the second case,$E_2 = 9\phi$,so $K.E_2 = 9\phi - \phi = 8\phi$.
Since $K.E = \frac{1}{2}mv^2$,we have the ratio $\frac{K.E_1}{K.E_2} = \frac{v_1^2}{v_2^2}$.
Substituting the values,$\frac{2\phi}{8\phi} = \frac{1}{4} = \frac{v_1^2}{v_2^2}$.
Taking the square root,$\frac{v_1}{v_2} = \frac{1}{2}$,which implies $v_2 = 2v_1$.
Given $v_1 = 6 \times 10^6 \ m/s$,then $v_2 = 2 \times 6 \times 10^6 = 12 \times 10^6 \ m/s$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K.E. = h n - \phi$,where $\phi$ is the work function of the metal.
For the first photocathode: $\frac{1}{2}mV_1^2 = hn_1 - \phi$ $(1)$
For the second photocathode: $\frac{1}{2}mV_2^2 = hn_2 - \phi$ $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$\frac{1}{2}mV_1^2 - \frac{1}{2}mV_2^2 = (hn_1 - \phi) - (hn_2 - \phi)$
$\frac{1}{2}m(V_1^2 - V_2^2) = h(n_1 - n_2)$
$V_1^2 - V_2^2 = \frac{2h}{m}(n_1 - n_2)$

(A) According to Einstein's photoelectric equation,the stopping potential $V_0$ is related to the frequency $f$ of the incident radiation by the equation: $eV_0 = hf - \Phi$,where $e$ is the charge of an electron,$h$ is Planck's constant,and $\Phi$ is the work function of the metal.
From the graph,the stopping potentials are $|V_0^1| < |V_0^2| < |V_0^3| < |V_0^4|$.
Since the stopping potential $V_0$ is directly proportional to the frequency $f$ for a given metal,a larger magnitude of stopping potential corresponds to a higher frequency of incident radiation.
Therefore,the order of frequencies is $f_a > f_b > f_c > f_d$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by: $K_{max} = hv - \phi$,where $h$ is Planck's constant,$v$ is the frequency of incident radiation,and $\phi$ is the work function of the metal.
Comparing this equation with the equation of a straight line,$y = mx + c$,we have $y = K_{max}$,$x = v$,slope $m = h$,and intercept $c = -\phi$.
Since the slope $h$ is positive and the intercept $-\phi$ is negative,the graph is a straight line starting from a threshold frequency $v_0$ (where $K_{max} = 0$) and increasing linearly with frequency $v$.
This corresponds to graph $(1)$.

(C) According to Einstein's photoelectric equation,the kinetic energy $E$ of an emitted photoelectron is given by $E = \frac{hc}{\lambda} - W_0$,where $W_0$ is the work function.
For wavelength $\lambda_1$,$E_1 = \frac{hc}{\lambda_1} - W_0 \implies E_1 \lambda_1 = hc - W_0 \lambda_1 \implies hc = E_1 \lambda_1 + W_0 \lambda_1$ ... $(i)$
For wavelength $\lambda_2$,$E_2 = \frac{hc}{\lambda_2} - W_0 \implies E_2 \lambda_2 = hc - W_0 \lambda_2 \implies hc = E_2 \lambda_2 + W_0 \lambda_2$ ... (ii)
Equating $(i)$ and (ii):
$E_1 \lambda_1 + W_0 \lambda_1 = E_2 \lambda_2 + W_0 \lambda_2$
$E_1 \lambda_1 - E_2 \lambda_2 = W_0 \lambda_2 - W_0 \lambda_1$
$E_1 \lambda_1 - E_2 \lambda_2 = W_0 (\lambda_2 - \lambda_1)$
$W_0 = \frac{E_1 \lambda_1 - E_2 \lambda_2}{\lambda_2 - \lambda_1}$

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $E_k$ is given by $E_k = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
For the first case with wavelength $\lambda$:
$E_1 = \frac{hc}{\lambda} - \phi$ ... $(i)$
For the second case with wavelength $\frac{\lambda}{2}$:
$E_2 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$ ... (ii)
Given that $E_1 = \frac{1}{4} E_2$,which implies $4E_1 = E_2$ ... (iii)
Substituting $(i)$ and (ii) into (iii):
$4\left(\frac{hc}{\lambda} - \phi\right) = \frac{2hc}{\lambda} - \phi$
$\frac{4hc}{\lambda} - 4\phi = \frac{2hc}{\lambda} - \phi$
$\frac{4hc}{\lambda} - \frac{2hc}{\lambda} = 4\phi - \phi$
$\frac{2hc}{\lambda} = 3\phi$
$\phi = \frac{2hc}{3\lambda}$

(A) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = hv - W_0$,where $h$ is Planck's constant,$v$ is the frequency of incident radiation,and $W_0$ is the work function of the metal.
Since the stopping potential $V_s$ is related to the maximum kinetic energy by $eV_s = K_{max}$,we have $eV_s = hv - W_0$.
Rearranging for $V_s$,we get $V_s = \frac{h}{e}v - \frac{W_0}{e}$.
From this equation,it is clear that the stopping potential $V_s$ is directly proportional to the frequency $v$ of the incident radiation.
Therefore,if the frequency $v$ is increased,the stopping potential $V_s$ will also increase.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
For the first case with wavelength $\lambda$:
$K_1 = \frac{hc}{\lambda} - \phi$ ... $(i)$
For the second case with wavelength $\lambda / 3$:
$K_2 = \frac{hc}{\lambda / 3} - \phi = \frac{3hc}{\lambda} - \phi$ ... $(ii)$
Given that $K_2 = 4K_1$,we substitute the expressions:
$\frac{3hc}{\lambda} - \phi = 4 \left( \frac{hc}{\lambda} - \phi \right)$
Expanding the equation:
$\frac{3hc}{\lambda} - \phi = \frac{4hc}{\lambda} - 4\phi$
Rearranging to solve for $\phi$:
$4\phi - \phi = \frac{4hc}{\lambda} - \frac{3hc}{\lambda}$
$3\phi = \frac{hc}{\lambda}$
$\phi = \frac{hc}{3\lambda}$

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ is given by: $K_{max} = h\nu - \Phi_0$,where $\Phi_0$ is the work function.
Since $K_{max} = eV_0$,we can write: $eV_0 = h\nu - \Phi_0$.
Dividing by $e$,we get: $V_0 = (\frac{h}{e})\nu - \frac{\Phi_0}{e}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = V_0$ and $x = \nu$,the slope $(m)$ is $\frac{h}{e}$.
Multiplying the slope by the charge of the electron $(e)$,we get: $m \times e = (\frac{h}{e}) \times e = h$.
Thus,the product gives the Planck's constant $(h)$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K.E_{\max} = E - \phi_0$,where $E$ is the energy of the incident photon and $\phi_0$ is the work function.
For the first case,$E_1 = 2\phi_0$,so $K.E_1 = 2\phi_0 - \phi_0 = \phi_0$.
For the second case,$E_2 = 3\phi_0$,so $K.E_2 = 3\phi_0 - \phi_0 = 2\phi_0$.
Since $K.E. = \frac{1}{2}mv^2$,we have $\frac{K.E_1}{K.E_2} = \frac{v_1^2}{v_2^2}$.
Substituting the values,$\frac{v_1^2}{v_2^2} = \frac{\phi_0}{2\phi_0} = \frac{1}{2}$.
Therefore,the ratio of maximum velocities is $\frac{v_1}{v_2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

(B) According to Einstein's photoelectric equation, the stopping potential $V_s$ is given by:
$e V_s = h \nu - \Phi_0$
$V_s = \frac{h}{e} \nu - \frac{\Phi_0}{e}$
where $\Phi_0$ is the work function and $\nu_0$ is the threshold frequency, such that $\Phi_0 = h \nu_0$.
From the given graph, the intercept on the frequency axis represents the threshold frequency $\nu_0$.
Since the threshold frequency $\nu_{0,A}$ for surface $A$ is less than the threshold frequency $\nu_{0,B}$ for surface $B$ ( $\nu_{0,A} < \nu_{0,B}$ ),
it follows that the work function $\Phi_{0,A} = h \nu_{0,A}$ is smaller than the work function $\Phi_{0,B} = h \nu_{0,B}$.
Therefore, the work function of $A$ is smaller than that of $B$.

(A) Using Einstein's photoelectric equation,$h \nu = \phi_0 + KE_{\text{max}}$.
Since $KE_{\text{max}} = eV_s$,we have $\frac{hc}{\lambda} = \phi_0 + e(4V_0)$ ....$(i)$
Similarly,for wavelength $3\lambda$,we have $\frac{hc}{3\lambda} = \phi_0 + eV_0$ ....(ii)
Multiply equation (ii) by $4$:
$\frac{4hc}{3\lambda} = 4\phi_0 + 4eV_0$ ....(iii)
Subtract equation $(i)$ from equation (iii):
$\frac{4hc}{3\lambda} - \frac{hc}{\lambda} = (4\phi_0 + 4eV_0) - (\phi_0 + 4eV_0)$
$\frac{hc}{3\lambda} = 3\phi_0$
$\phi_0 = \frac{hc}{9\lambda}$
Since the work function $\phi_0 = \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength,we get $\lambda_0 = 9\lambda$.
Thus,option $(A)$ is correct.

(C) The stopping potential $(V_0)$ in the photoelectric effect is determined by the maximum kinetic energy of the emitted photoelectrons, which depends solely on the frequency of the incident light and the work function of the metal surface, as given by Einstein's photoelectric equation: $eV_0 = h\nu - \Phi$.
Moving the point source of light farther away changes the intensity of the light incident on the metal surface, but it does not change the frequency of the incident photons.
Since the stopping potential is independent of the intensity of the incident light, it will remain constant.

(B) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by $eV_0 = h\nu - h\nu_0$,where $\nu_0$ is the threshold frequency.
For frequency $\nu$,we have: $eV_0 = h\nu - h\nu_0$ ... $(i)$
For frequency $\frac{\nu}{2}$,we have: $e\frac{V_0}{4} = h\frac{\nu}{2} - h\nu_0$ ... $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$4 = \frac{h\nu - h\nu_0}{\frac{h\nu}{2} - h\nu_0} = \frac{\nu - \nu_0}{\frac{\nu}{2} - \nu_0}$
$4(\frac{\nu}{2} - \nu_0) = \nu - \nu_0$
$2\nu - 4\nu_0 = \nu - \nu_0$
$2\nu - \nu = 4\nu_0 - \nu_0$
$\nu = 3\nu_0$
$\nu_0 = \frac{\nu}{3}$

(B) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K.E_{\max} = E - \phi_0$,where $E$ is the energy of the incident photon and $\phi_0$ is the work function of the metal.
For the first case,$E_1 = 2\phi_0$. Therefore,$K.E_1 = 2\phi_0 - \phi_0 = \phi_0$.
Since $K.E_1 = \frac{1}{2}mv_1^2$,we have $\frac{1}{2}mv_1^2 = \phi_0$ --- $(1)$
For the second case,$E_2 = 3\phi_0$. Therefore,$K.E_2 = 3\phi_0 - \phi_0 = 2\phi_0$.
Since $K.E_2 = \frac{1}{2}mv_2^2$,we have $\frac{1}{2}mv_2^2 = 2\phi_0$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{\phi_0}{2\phi_0}$
$\frac{v_1^2}{v_2^2} = \frac{1}{2}$
Taking the square root on both sides:
$\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}$
Thus,the ratio $v_1: v_2$ is $1: \sqrt{2}$.

(B) According to Einstein's photoelectric equation: $E = \frac{hc}{\lambda} - \phi_0$ ... $(i)$
Given that when the wavelength becomes $\lambda' = \frac{\lambda}{3}$,the kinetic energy becomes $E' = 4E$.
Substituting these into the photoelectric equation: $4E = \frac{hc}{\lambda/3} - \phi_0$
$4E = \frac{3hc}{\lambda} - \phi_0$ ... (ii)
Substitute the value of $E$ from equation $(i)$ into equation (ii):
$4\left(\frac{hc}{\lambda} - \phi_0\right) = \frac{3hc}{\lambda} - \phi_0$
$\frac{4hc}{\lambda} - 4\phi_0 = \frac{3hc}{\lambda} - \phi_0$
$\frac{4hc}{\lambda} - \frac{3hc}{\lambda} = 4\phi_0 - \phi_0$
$\frac{hc}{\lambda} = 3\phi_0$
$\phi_0 = \frac{hc}{3\lambda}$

(D) According to Einstein's photoelectric equation, the maximum kinetic energy $(K_{max})$ of photoelectrons is given by:
$K_{max} = h\nu - \Phi_0$
where $h$ is Planck's constant, $\nu$ is the frequency of incident radiation, and $\Phi_0$ is the work function of the metal.
From this equation, it is clear that $K_{max}$ varies linearly with the frequency $(\nu)$ of the incident radiation.
Furthermore, the maximum kinetic energy is independent of the intensity of the incident radiation, as intensity only affects the number of photoelectrons emitted per unit time.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function of the metal.
For wavelength $\lambda_1$,$K_1 = \frac{hc}{\lambda_1} - \phi$ $(i)$
For wavelength $\lambda_2$,$K_2 = \frac{hc}{\lambda_2} - \phi$ (ii)
Given $\lambda_1 = 3 \lambda_2$,substitute this into equation $(i)$:
$K_1 = \frac{hc}{3 \lambda_2} - \phi$ (iii)
From equation (ii),$\frac{hc}{\lambda_2} = K_2 + \phi$. Substituting this into equation (iii):
$K_1 = \frac{1}{3} (K_2 + \phi) - \phi$
$K_1 = \frac{K_2}{3} + \frac{\phi}{3} - \phi$
$K_1 = \frac{K_2}{3} - \frac{2\phi}{3}$
Since the work function $\phi > 0$,it follows that $K_1 < \frac{K_2}{3}$.

(B) Energy of incident light $(E)$:
$E = \frac{hc}{\lambda} = \frac{1240 eV \cdot nm}{310 nm} = 4 eV$
According to Einstein's photoelectric equation:
$K_{max} = E - \phi_0$
Where $\phi_0$ is the work function.
$K_{max} = 4 eV - 1.13 eV = 2.87 eV$
The stopping potential $(V_0)$ is given by $K_{max} = eV_0$.
Therefore, $V_0 = 2.87 V$.

(D) The photoelectric equation is given by $eV_0 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$,where $\lambda_0$ is the threshold wavelength.
For the first case: $eV = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \dots (i)$
For the second case: $e \left( \frac{V}{6} \right) = hc \left( \frac{1}{3\lambda} - \frac{1}{\lambda_0} \right) \dots (ii)$
Dividing equation $(i)$ by equation $(ii)$:
$6 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{3\lambda} - \frac{1}{\lambda_0}} = \frac{\frac{\lambda_0 - \lambda}{\lambda \lambda_0}}{\frac{\lambda_0 - 3\lambda}{3\lambda \lambda_0}} = \frac{3(\lambda_0 - \lambda)}{\lambda_0 - 3\lambda}$
$6(\lambda_0 - 3\lambda) = 3\lambda_0 - 3\lambda$
$6\lambda_0 - 18\lambda = 3\lambda_0 - 3\lambda$
$3\lambda_0 = 15\lambda$
$\lambda_0 = 5\lambda$

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E._{\max})$ is given by $K.E._{\max} = E - \phi_0$,where $E$ is the energy of the incident photon and $\phi_0$ is the work function of the metal.
For the first case,$E_1 = 2\phi_0$,so $K.E._{\max 1} = 2\phi_0 - \phi_0 = \phi_0$.
For the second case,$E_2 = 5\phi_0$,so $K.E._{\max 2} = 5\phi_0 - \phi_0 = 4\phi_0$.
The ratio of maximum kinetic energies is $\frac{K.E._{\max 1}}{K.E._{\max 2}} = \frac{\phi_0}{4\phi_0} = \frac{1}{4}$.
Since $K.E._{\max} = \frac{1}{2}mv_{\max}^2$,the ratio of velocities is $\frac{v_1}{v_2} = \sqrt{\frac{K.E._{\max 1}}{K.E._{\max 2}}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.

(B) The work function $\phi_0$ of a photoelectric emitter is given by the formula $\phi_0 = \frac{hc}{\lambda_0}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda_0$ is the threshold wavelength.
From this relation,we can see that $\phi_0 \propto \frac{1}{\lambda_0}$.
For the first emitter,the incident wavelength is $300 \ nm$. Since photoelectrons are emitted,the threshold wavelength $\lambda_{0_1}$ must be at least $300 \ nm$.
For the second emitter,the threshold wavelength $\lambda_{0_2}$ is given as $600 \ nm$.
Assuming the first emitter is just at the threshold for $300 \ nm$,we have $\lambda_{0_1} = 300 \ nm$ and $\lambda_{0_2} = 600 \ nm$.
The ratio of the work functions is $\frac{\phi_{0_1}}{\phi_{0_2}} = \frac{\lambda_{0_2}}{\lambda_{0_1}} = \frac{600 \ nm}{300 \ nm} = \frac{2}{1}$.
Thus,the ratio is $2: 1$.

(B) 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.
For the first case: $4.8 = \frac{hc}{e\lambda} - \frac{\phi}{e} \quad \dots(i)$
For the second case,where wavelength is $2\lambda$: $1.6 = \frac{hc}{e(2\lambda)} - \frac{\phi}{e} \quad \dots(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$4.8 - 1.6 = \left(\frac{hc}{e\lambda} - \frac{\phi}{e}\right) - \left(\frac{hc}{2e\lambda} - \frac{\phi}{e}\right)$
$3.2 = \frac{hc}{e\lambda} - \frac{hc}{2e\lambda} = \frac{hc}{2e\lambda}$
So,$\frac{hc}{e\lambda} = 6.4$.
Substituting this into equation $(i)$:
$4.8 = 6.4 - \frac{\phi}{e} \Rightarrow \frac{\phi}{e} = 6.4 - 4.8 = 1.6$.
The threshold wavelength $\lambda_0$ is given by $\phi = \frac{hc}{\lambda_0}$,so $\frac{\phi}{e} = \frac{hc}{e\lambda_0} = 1.6$.
Since $\frac{hc}{e\lambda} = 6.4$,we have $\frac{hc}{e\lambda_0} = \frac{1}{4} \left(\frac{hc}{e\lambda}\right)$.
Therefore,$\lambda_0 = 4\lambda$.

(B) Concept: Photoelectric effect and the motion of a charge in a direction perpendicular to a magnetic field.
According to Einstein's photoelectric equation,the maximum kinetic energy $K$ of a photoelectron is given by:
$K = E - W_0$
Since $K = \frac{1}{2} m v^2$,we have:
$v = \sqrt{\frac{2(E - W_0)}{m}}$
When a charge $e$ moves with velocity $v$ perpendicular to a magnetic field $B$,it experiences a magnetic Lorentz force that acts as a centripetal force,causing it to move in a circular path of radius $r$:
$e v B = \frac{m v^2}{r}$
Solving for $r$:
$r = \frac{m v}{e B}$
Substituting the expression for $v$:
$r = \frac{m}{e B} \sqrt{\frac{2(E - W_0)}{m}} = \frac{\sqrt{m^2 \cdot \frac{2(E - W_0)}{m}}}{e B} = \frac{\sqrt{2m(E - W_0)}}{e B}$

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
For the first case with wavelength $\lambda$: $K_1 = \frac{hc}{\lambda} - \phi$.
For the second case with wavelength $\frac{\lambda}{2}$: $K_2 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$.
Given that $K_1 = \frac{1}{3} K_2$,we have $3K_1 = K_2$.
Substituting the expressions: $3(\frac{hc}{\lambda} - \phi) = \frac{2hc}{\lambda} - \phi$.
$\frac{3hc}{\lambda} - 3\phi = \frac{2hc}{\lambda} - \phi$.
$\frac{3hc}{\lambda} - \frac{2hc}{\lambda} = 3\phi - \phi$.
$\frac{hc}{\lambda} = 2\phi$.
Therefore,$\phi = \frac{hc}{2\lambda}$.