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

(B) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = \frac{1}{2} m v^2 = E_p - \phi$,where $E_p$ is the energy of the incident photon and $\phi$ is the work function of the metal.
Given $\phi = 0.8 eV$.
For the first photon with energy $E_1 = 1.3 eV$:
$\frac{1}{2} m v_1^2 = 1.3 eV - 0.8 eV = 0.5 eV$ --- $(1)$
For the second photon with energy $E_2 = 2.8 eV$:
$\frac{1}{2} m v_2^2 = 2.8 eV - 0.8 eV = 2.0 eV$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{0.5 eV}{2.0 eV}$
$\frac{v_1^2}{v_2^2} = \frac{0.5}{2.0} = \frac{1}{4}$
Taking the square root on both sides:
$\frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
Therefore,the ratio of maximum speeds is $1: 2$.

(C) Concept: The stopping potential $V$ is related to the maximum kinetic energy $K_{\max}$ of the photoelectrons via Einstein's photoelectric effect equation:
$eV = K_{\max} = hv - hv_0$
Therefore,
$V = \left(\frac{h}{e}\right)v - \left(\frac{h}{e}\right)v_0$
This equation is of the form $y = mx + c$,which represents a straight line.
Here,the slope $m = \frac{h}{e}$ is positive,and the intercept on the $V$-axis is $-\left(\frac{h}{e}\right)v_0$.
At $v = v_0$,the stopping potential $V = 0$.
Thus,the graph is a straight line starting from $v = v_0$ on the frequency axis and increasing linearly with frequency.
Graph $(A)$ correctly represents this relationship.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E._{max})$ of emitted electrons is given by:
$K.E._{max} = E - \phi$
Where $E$ is the energy of the incident photon and $\phi$ is the work function of the metal.
Given that the cutoff frequency (threshold frequency) is $v$,the work function is $\phi = hv$.
The energy of the incident radiation with frequency $2v$ is $E = h(2v) = 2hv$.
Substituting these values into the equation:
$\frac{1}{2}mv_{max}^2 = 2hv - hv$
$\frac{1}{2}mv_{max}^2 = hv$
$v_{max}^2 = \frac{2hv}{m}$
$v_{max} = \sqrt{\frac{2hv}{m}}$

(B) Einstein's photoelectric equation is given by: $e V = \frac{h c}{\lambda} - \phi$,which can be written as $\frac{h c}{\lambda} = \phi + e V$.
For incident wavelength $\lambda$,the stopping potential is $V_1$:
$\frac{h c}{\lambda} = \phi + e V_1$ --- $(1)$
For incident wavelength $\frac{\lambda}{2}$,the stopping potential is $V_2$:
$\frac{h c}{\lambda / 2} = \phi + e V_2 \Rightarrow \frac{2 h c}{\lambda} = \phi + e V_2$ --- $(2)$
Substitute the value of $\frac{h c}{\lambda}$ from equation $(1)$ into equation $(2)$:
$2(\phi + e V_1) = \phi + e V_2$
$2 \phi + 2 e V_1 = \phi + e V_2$
$e V_2 = 2 e V_1 + \phi$
Dividing by $e$,we get:
$V_2 = 2 V_1 + \frac{\phi}{e}$

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = h v - h v_0$,where $v_0$ is the threshold frequency.
Since the stopping potential $V_s$ is related to the maximum kinetic energy by $K_{max} = e V_s$,we have $e V_s = h v - h v_0$ ... $(1)$.
For the second case,with frequency $v/2$ and stopping potential $V_s/4$,we have $e(V_s/4) = h(v/2) - h v_0$ ... $(2)$.
From equation $(1)$,$e V_s = h v - h v_0$. Substituting this into equation $(2)$ multiplied by $4$:
$e V_s = 4(h v/2 - h v_0) = 2h v - 4h v_0$.
Equating the two expressions for $e V_s$:
$h v - h v_0 = 2h v - 4h v_0$.
Rearranging the terms: $3h v_0 = h v$.
Therefore,$v_0 = v/3$.

(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Given $h = 6.63 \times 10^{-34} J s$, $c = 3 \times 10^8 m/s$, and $\lambda = 4100 \times 10^{-10} m$.
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4100 \times 10^{-10}} J = 4.85 \times 10^{-19} J$.
To convert this energy into electron-volts $(eV)$, divide by $e = 1.6 \times 10^{-19} C$:
$E = \frac{4.85 \times 10^{-19}}{1.6 \times 10^{-19}} eV \approx 3.03 eV$.
Photoelectric emission occurs if the incident photon energy is greater than the work function $(\Phi)$ of the metal.
For metal $A$: $\Phi_A = 1.92 eV < 3.03 eV$ (Emission occurs).
For metal $B$: $\Phi_B = 2.0 eV < 3.03 eV$ (Emission occurs).
For metal $C$: $\Phi_C = 5 eV > 3.03 eV$ (No emission).
Therefore, metals $A$ and $B$ will emit photoelectrons.

(A) According to Einstein's photoelectric equation:
$eV = \frac{hc}{\lambda} - \Phi$ --- $(1)$
where $V$ is the stopping potential,$\lambda$ is the incident wavelength,and $\Phi$ is the work function of the metal.
When the incident wavelength is changed to $2\lambda$,let the new stopping potential be $V'$. The equation becomes:
$eV' = \frac{hc}{2\lambda} - \Phi$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$eV - eV' = \left( \frac{hc}{\lambda} - \Phi \right) - \left( \frac{hc}{2\lambda} - \Phi \right)$
$e(V - V') = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda}$
$V - V' = \frac{hc}{2e\lambda}$
$V' = V - \frac{hc}{2e\lambda}$

(A) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ of the emitted photoelectron is given by:
$K_{max} = h\nu - \phi$
where $\phi$ is the work function of the metal.
Given that the threshold frequency is $\nu$, the work function is $\phi = h\nu$.
When radiation of frequency $2\nu$ is incident on the metal, the maximum kinetic energy is:
$K_{max} = h(2\nu) - h\nu = h\nu$
Since $K_{max} = \frac{1}{2}mv_{max}^2$, we have:
$\frac{1}{2}mv_{max}^2 = h\nu$
$v_{max}^2 = \frac{2h\nu}{m}$
$v_{max} = \sqrt{\frac{2h\nu}{m}}$
Therefore, option $A$ is correct.

(B) The kinetic energy of the fastest emitted photoelectron is given by $K_{max} = \frac{1}{2} m v^2$.
The stopping potential $V$ is the potential required to stop these fastest electrons,such that the work done by the electric field equals the maximum kinetic energy:
$e V = K_{max}$
$e V = \frac{1}{2} m v^2$
Solving for $V$,we get:
$V = \frac{m v^2}{2 e}$

(C) According to Einstein's photoelectric equation, the maximum kinetic energy $(K_{max})$ of an emitted photoelectron is given by $K_{max} = h\nu - \Phi$, where $h$ is Planck's constant, $\nu$ is the frequency of the incident light, and $\Phi$ is the work function of the metal surface.
Since $h$ and $\Phi$ are constants for a given metal, the kinetic energy $K_{max}$ depends directly on the frequency $\nu$ of the incident light.
Intensity of light affects the number of photoelectrons emitted per second, but not their individual kinetic energy.

(B) The maximum kinetic energy of the emitted photoelectrons is given by Einstein's photoelectric equation:
$(KE)_{\max} = h\nu - \phi = 4.2 \text{ eV} - 2.4 \text{ eV} = 1.8 \text{ eV}$.
Emission stops when the potential $V$ of the sphere reaches a value such that the potential energy of an electron equals its maximum kinetic energy:
$eV = (KE)_{\max} \implies V = 1.8 \text{ V}$.
The potential of a metallic sphere of radius $r$ is given by $V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}$.
Substituting the values: $1.8 = (9 \times 10^9) \times \frac{q}{0.1}$.
Solving for charge $q$: $q = \frac{1.8 \times 0.1}{9 \times 10^9} = 2 \times 10^{-11} \text{ C}$.
The number of photoelectrons $n$ emitted is $n = \frac{q}{e} = \frac{2 \times 10^{-11}}{1.6 \times 10^{-19}} = 1.25 \times 10^8$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E.)_{max}$ is given by:
$(K.E.)_{max} = h v - h v_0$
where $h$ is Planck's constant,$v$ is the frequency of incident light,and $v_0$ is the threshold frequency.
For frequencies $v_1$ and $v_2$,we have:
$(K.E.)_1 = h v_1 - h v_0$ ....$(1)$
$(K.E.)_2 = h v_2 - h v_0$ ....$(2)$
Given the ratio of maximum kinetic energies is $\frac{(K.E.)_1}{(K.E.)_2} = \frac{1}{K}$.
Dividing equation $(1)$ by $(2)$:
$\frac{1}{K} = \frac{h v_1 - h v_0}{h v_2 - h v_0} = \frac{v_1 - v_0}{v_2 - v_0}$
Cross-multiplying:
$v_2 - v_0 = K(v_1 - v_0)$
$v_2 - v_0 = K v_1 - K v_0$
$K v_0 - v_0 = K v_1 - v_2$
$v_0(K - 1) = K v_1 - v_2$
$v_0 = \frac{K v_1 - v_2}{K - 1}$

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E.)_{max}$ is given by:
$(K.E.)_{max} = \frac{hc}{\lambda} - W_0$,where $W_0$ is the work function.
For the first case:
$(K.E.)_1 = \frac{1}{2}mV^2 = \frac{hc}{\lambda} - W_0$ ... $(1)$
For the second case,with wavelength $\lambda' = \frac{2\lambda}{3}$:
$(K.E.)_2 = \frac{1}{2}mv_2^2 = \frac{hc}{\lambda'} - W_0 = \frac{hc}{(2\lambda/3)} - W_0 = \frac{3}{2}\frac{hc}{\lambda} - W_0$ ... $(2)$
From equation $(1)$,$\frac{hc}{\lambda} = \frac{1}{2}mV^2 + W_0$. Substituting this into equation $(2)$:
$(K.E.)_2 = \frac{3}{2}(\frac{1}{2}mV^2 + W_0) - W_0$
$(K.E.)_2 = \frac{3}{2}(\frac{1}{2}mV^2) + \frac{3}{2}W_0 - W_0$
$(K.E.)_2 = \frac{3}{2}(\frac{1}{2}mV^2) + \frac{1}{2}W_0$
Since $W_0 > 0$,it follows that $(K.E.)_2 > \frac{3}{2}(K.E.)_1$.
$\frac{1}{2}mv_2^2 > \frac{3}{2}(\frac{1}{2}mV^2)$
$v_2^2 > 1.5 V^2$
$v_2 > \sqrt{1.5} V = (1.5)^{1/2} V$.

(D) The energy of the incident photons is $E = 10 eV$.
The threshold frequency is $\nu_0 = 2 \times 10^{15} Hz$.
The work function $W_0$ is given by $W_0 = h\nu_0$.
Substituting the values: $W_0 = (6.63 \times 10^{-34} Js) \times (2 \times 10^{15} Hz) = 13.26 \times 10^{-19} J$.
To convert this into $eV$,divide by the charge of an electron $(1.6 \times 10^{-19} C)$:
$W_0 = \frac{13.26 \times 10^{-19}}{1.6 \times 10^{-19}} eV \approx 8.2875 eV \approx 8.3 eV$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - W_0$.
$K_{max} = 10 eV - 8.3 eV = 1.7 eV$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of the emitted electron is given by:
$K_{\max} = E - \phi$
where $E = \frac{hc}{\lambda}$ is the energy of the incident photon.
Substituting the values,we get:
$\frac{1}{2} mv_{\max}^2 = \frac{hc}{\lambda} - \phi$
$\frac{1}{2} mv_{\max}^2 = \frac{hc - \phi \lambda}{\lambda}$
Multiplying both sides by $\frac{2}{m}$:
$v_{\max}^2 = \frac{2(hc - \phi \lambda)}{m \lambda}$
Taking the square root on both sides:
$v_{\max} = \sqrt{\frac{2(hc - \phi \lambda)}{m \lambda}}$
Thus,the correct option is $D$.

(B) According to Einstein's photoelectric equation,the kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - W_0$,where $W_0$ is the work function.
For wavelength $\lambda_1$,the kinetic energy is $K_1 = \frac{hc}{\lambda_1} - W_0$.
For wavelength $\lambda_2$,the kinetic energy is $K_2 = \frac{hc}{\lambda_2} - W_0$.
Given that $K_2 = 3K_1$,we substitute the expressions:
$\frac{hc}{\lambda_2} - W_0 = 3\left(\frac{hc}{\lambda_1} - W_0\right)$.
$\frac{hc}{\lambda_2} - W_0 = \frac{3hc}{\lambda_1} - 3W_0$.
Rearranging the terms to solve for $W_0$:
$3W_0 - W_0 = \frac{3hc}{\lambda_1} - \frac{hc}{\lambda_2}$.
$2W_0 = hc\left(\frac{3}{\lambda_1} - \frac{1}{\lambda_2}\right)$.
$2W_0 = hc\left(\frac{3\lambda_2 - \lambda_1}{\lambda_1\lambda_2}\right)$.
Therefore,$W_0 = \frac{hc}{2}\left(\frac{3\lambda_2 - \lambda_1}{\lambda_1\lambda_2}\right)$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy is given by:
$(KE)_{\max} = h\nu - \phi_0$
where $\phi_0 = h\nu_0$ is the work function.
Comparing this with the equation of a straight line $y = mx + c$:
Here,$y = (KE)_{\max}$,$x = \nu$,$m = h$ (slope),and $c = -h\nu_0$ (y-intercept).
$1$. The $X$-intercept $(A)$ occurs when $(KE)_{\max} = 0$:
$0 = h\nu - h\nu_0 \implies \nu = \nu_0 = A$.
$2$. The $Y$-intercept $(B)$ occurs when $\nu = 0$:
$(KE)_{\max} = -h\nu_0 = B$. Since $B$ represents the magnitude of the intercept,we take $|B| = h\nu_0$.
Therefore,the ratio of the magnitude of the $Y$-intercept to the $X$-intercept is:
$\frac{|B|}{A} = \frac{h\nu_0}{\nu_0} = h$.
Thus,Planck's constant $h$ is given by $\frac{B}{A}$ (considering magnitudes).

(D) According to Einstein's photoelectric equation,the kinetic energy $E$ of an emitted photoelectron is given by $E = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function of the material. This shows that $E$ depends only on the wavelength $\lambda$ of the incident light,specifically $E \propto \frac{1}{\lambda}$.
Intensity $I$ of light is proportional to the number of photons incident per unit area per unit time. Since each photon interacts with one electron to cause emission,the number of emitted photoelectrons $N$ is directly proportional to the intensity $I$ of the incident light $(N \propto I)$.

(D) The kinetic energy of the emitted photoelectron is given by $K_{max} = \frac{1}{2}mv^2$.
At the stopping potential $V_s$,the work done by the retarding electric field is equal to the maximum kinetic energy of the photoelectron.
Therefore,$eV_s = \frac{1}{2}mv^2$.
Solving for the stopping potential $V_s$:
$V_s = \frac{mv^2}{2e}$.
This can be rewritten as $V_s = \frac{v^2}{2(e/m)}$.

(A) 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 metal.
Initially, $K_{max1} = h\nu - \phi$.
When the frequency is doubled, the new frequency is $\nu' = 2\nu$.
The new maximum kinetic energy is $K_{max2} = h(2\nu) - \phi = 2h\nu - \phi$.
Since $K_{max2} = 2(h\nu - \phi/2)$, and $\phi > 0$, it follows that $K_{max2} > 2(h\nu - \phi) = 2K_{max1}$.
Therefore, the maximum kinetic energy increases to slightly more than double the initial value.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E.)_{\max}$ of emitted photoelectrons is given by:
$(K.E.)_{\max} = h\nu - \phi_0$
Where:
$h\nu$ is the energy of the incident photon.
$\phi_0$ is the work function of the metal.
From the equation,it is clear that $(K.E.)_{\max}$ is inversely proportional to the work function $\phi_0$ for a constant incident frequency $\nu$.
Therefore,if the work function $\phi_0$ increases,the maximum kinetic energy $(K.E.)_{\max}$ decreases.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy '$K$' of the emitted photoelectron is given by: $K = E - W_{0}$.
Since $K = \frac{P^{2}}{2m}$,the momentum '$P$' of the electron is $P = \sqrt{2mK} = \sqrt{2m(E - W_{0})}$.
When a charged particle moves in a uniform magnetic field '$B$' perpendicular to its velocity,it follows a circular path of radius '$r = \frac{P}{eB}$.
Substituting the value of '$P$',we get: $r = \frac{\sqrt{2m(E - W_{0})}}{eB}$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(E_k)$ of emitted photoelectrons is given by:
$E_k = h\nu - \Phi$
Where $h$ is Planck's constant,$\nu$ is the frequency of incident radiation,and $\Phi$ is the work function of the metal.
This equation is of the form $y = mx + c$,where $y = E_k$,$x = \nu$,$m = h$ (slope),and $c = -\Phi$ (y-intercept).
Since the y-intercept is negative $(-\Phi)$,the graph is a straight line that does not pass through the origin. It starts from the x-axis at a threshold frequency $\nu_0$ (where $E_k = 0$,so $h\nu_0 = \Phi$) and has a positive slope $h$. The provided image shows a straight line with a positive x-intercept,which correctly represents this relationship.

(A) 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 radiation,and $\Phi$ is the work function of the metal surface.
This equation shows that the maximum kinetic energy depends only on the frequency of the incident radiation and the work function of the metal. It is independent of the intensity $(I)$ of the incident radiation.
Therefore,the graph showing the variation of maximum kinetic energy $(E)$ with intensity $(I)$ should be a horizontal line parallel to the intensity axis. Assuming graph $(D)$ represents this constant behavior,the correct option is $(D)$.

(B) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
Since $K_{max} = eV_{s}$,where $V_{s}$ is the stopping potential,we have $eV_{s} = \frac{hc}{\lambda} - \phi$.
For the first case: $eV_{0} = \frac{hc}{\lambda_{0}} - \phi$ --- $(1)$
For the second case with wavelength $2\lambda_{0}$: $eV_{s}' = \frac{hc}{2\lambda_{0}} - \phi$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$eV_{0} - eV_{s}' = \left(\frac{hc}{\lambda_{0}} - \phi\right) - \left(\frac{hc}{2\lambda_{0}} - \phi\right)$
$e(V_{0} - V_{s}') = \frac{hc}{\lambda_{0}} - \frac{hc}{2\lambda_{0}} = \frac{hc}{2\lambda_{0}}$
$V_{0} - V_{s}' = \frac{hc}{2e\lambda_{0}}$
$V_{s}' = V_{0} - \frac{hc}{2e\lambda_{0}}$
Note: If $\frac{hc}{2\lambda_{0}} < \phi$,the stopping potential will be $0$ as no photoelectric emission occurs.

(D) The threshold frequency of the material is $\nu_0$. The initial incident frequency is $\nu_1 = 2\nu_0$. Since $\nu_1 > \nu_0$,photoelectric emission occurs.
When the incident frequency is changed to $\nu_2 = \frac{1}{3} \nu_1 = \frac{1}{3} (2\nu_0) = \frac{2}{3} \nu_0$.
For photoelectric emission to occur,the incident frequency must be greater than or equal to the threshold frequency $(\nu \ge \nu_0)$.
Since $\nu_2 = \frac{2}{3} \nu_0 < \nu_0$,the incident frequency is now less than the threshold frequency.
Therefore,no photoelectric emission will take place regardless of the increase in intensity.
Thus,the photoelectric current will be zero.

(B) 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 wavelength $\lambda_{1}$,$E_{1} = \frac{hc}{\lambda_{1}} - \phi$ --- $(i)$
For wavelength $\lambda_{2}$,$E_{2} = \frac{hc}{\lambda_{2}} - \phi$ --- (ii)
From $(i)$,$hc = \lambda_{1}(E_{1} + \phi)$.
From (ii),$hc = \lambda_{2}(E_{2} + \phi)$.
Equating the two expressions for $hc$:
$\lambda_{1}(E_{1} + \phi) = \lambda_{2}(E_{2} + \phi)$
$\lambda_{1}E_{1} + \lambda_{1}\phi = \lambda_{2}E_{2} + \lambda_{2}\phi$
$\phi(\lambda_{1} - \lambda_{2}) = \lambda_{2}E_{2} - \lambda_{1}E_{1}$
$\phi = \frac{\lambda_{2}E_{2} - \lambda_{1}E_{1}}{\lambda_{1} - \lambda_{2}}$
Multiplying numerator and denominator by $-1$ gives $\phi = \frac{\lambda_{1}E_{1} - \lambda_{2}E_{2}}{\lambda_{2} - \lambda_{1}}$.

(B) The kinetic energy of the emitted photoelectron is given by $K_{max} = \frac{1}{2} mv^{2}$.
At the stopping potential $V_{s}$,the work done by the retarding potential is equal to the maximum kinetic energy,so $eV_{s} = \frac{1}{2} mv^{2}$.
Rearranging for $V_{s}$,we get $V_{s} = \frac{1}{2} \frac{m}{e} v^{2}$.
Since $\frac{m}{e} = \frac{1}{(e/m)}$,we can write $V_{s} = \frac{v^{2}}{2(e/m)}$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of an emitted photoelectron is given by $K_{max} = E - W$,where $E$ is the energy of the incident photon and $W$ is the work function.
For the first case,$E_1 = 3W$. Thus,$K_1 = \frac{1}{2}mv_1^2 = 3W - W = 2W$.
For the second case,$E_2 = 5W$. Thus,$K_2 = \frac{1}{2}mv_2^2 = 5W - W = 4W$.
Taking the ratio of the two equations:
$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{2W}{4W} = \frac{1}{2}$.
Therefore,$\frac{v_1^2}{v_2^2} = \frac{1}{2}$,which implies $\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}$.
The ratio of velocities is $1: \sqrt{2}$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E._{max})$ of emitted photoelectrons is given by:
$K.E._{max} = h\nu - \phi_0$
Comparing this with the equation of a straight line $y = mx + c$,where $y = K.E._{max}$ and $x = \nu$:
$1$. The slope $(m)$ of the graph is equal to Planck's constant $(h)$.
$2$. The $X$-axis intercept occurs when $K.E._{max} = 0$,which gives $0 = h\nu_0 - \phi_0$,or $\nu_0 = \phi_0 / h$. This intercept represents the threshold frequency $(\nu_0)$.
Therefore,the slope is Planck's constant and the $X$-axis intercept is the threshold frequency.

(B) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{\max}$ of emitted electrons is given by:
$K_{\max} = \frac{hc}{\lambda} - \phi$
Since $K_{\max} = \frac{1}{2}mv^2$, we have:
$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi$
$\frac{1}{2}mv^2 = \frac{hc - \lambda\phi}{\lambda}$
$v^2 = \frac{2(hc - \lambda\phi)}{m\lambda}$
$v = \left[\frac{2(hc - \lambda\phi)}{m\lambda}\right]^{1/2}$
Thus, the correct option is $B$.

(D) The work function $\phi$ of a metal is related to the threshold wavelength $\lambda_{0}$ by the formula $\phi = \frac{hc}{\lambda_{0}}$.
Given that $\lambda_{A} = 400 \ nm$ and $\lambda_{B} = 800 \ nm$.
Therefore,the work functions are $\phi_{A} = \frac{hc}{\lambda_{A}}$ and $\phi_{B} = \frac{hc}{\lambda_{B}}$.
The ratio $\frac{\phi_{A}}{\phi_{B}} = \frac{hc/\lambda_{A}}{hc/\lambda_{B}} = \frac{\lambda_{B}}{\lambda_{A}}$.
Substituting the values,we get $\frac{\phi_{A}}{\phi_{B}} = \frac{800 \ nm}{400 \ nm} = 2$.
Thus,the ratio $\phi_{A} : \phi_{B}$ is $2$.

(C) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by:
$eV_s = h\nu - \phi_o = h\nu - h\nu_o$
$V_s = \frac{h}{e}(\nu - \nu_o)$
Where $h$ is Planck's constant,$e$ is the charge of an electron,$\nu$ is the frequency of incident radiation,and $\nu_o$ is the threshold frequency.
For a fixed incident frequency $\nu$,the stopping potential $V_s$ is inversely proportional to the threshold frequency $\nu_o$.
From the graph,the threshold frequencies follow the order: $\nu_o(A) < \nu_o(B) < \nu_o(C) < \nu_o(D)$.
Since metal $A$ has the lowest threshold frequency,it will have the highest stopping potential for any given incident frequency $\nu$ (where $\nu > \nu_o(D)$).
Therefore,the correct option is $C$.

(C) According to Einstein's photoelectric equation,the stopping potential $V_{s}$ is given by:
$eV_{s} = h\nu - \phi$
$V_{s} = \frac{h}{e}\nu - \frac{\phi}{e}$
Comparing this with the equation of a straight line $y = mx + c$,the x-intercept (where $V_{s} = 0$) is the threshold frequency $\nu_{0}$,where $\nu_{0} = \frac{\phi}{h}$ or $\phi = h\nu_{0}$.
From the given graph,the threshold frequency for metal '$P$' is $\nu_{0}$ and for metal '$Q$' is $\nu_{0}^{\prime}$.
Since $\nu_{0} < \nu_{0}^{\prime}$,it follows that $h\nu_{0} < h\nu_{0}^{\prime}$.
Therefore,the work function $\phi_{P} < \phi_{Q}$.

(D) $1$. The stopping potential $V_s$ is related to the maximum kinetic energy $(K.E.)_{\max}$ by the equation: $(K.E.)_{\max} = e V_s$.
Given $(K.E.)_{\max} = 3.2 \times 10^{-19} \text{ J}$ and $e = 1.6 \times 10^{-19} \text{ C}$.
$V_s = \frac{(K.E.)_{\max}}{e} = \frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 2 \text{ V}$.
$2$. The threshold wavelength $\lambda_0$ is related to the work function $\Phi$ by the equation: $\Phi = \frac{hc}{\lambda_0}$.
Given $\Phi = 6.63 \times 10^{-19} \text{ J}$,$h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$,and $c = 3 \times 10^{8} \text{ m/s}$.
$\lambda_0 = \frac{hc}{\Phi} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{6.63 \times 10^{-19}} = 3 \times 10^{-7} \text{ m} = 3000 \times 10^{-10} \text{ m} = 3000 \text{ Å}$.
Thus,the stopping potential is $2 \text{ V}$ and the threshold wavelength is $3000 \text{ Å}$. The correct option is $(D)$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $(E_k)$ of emitted photoelectrons is given by $E_k = h\nu - \phi$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\phi$ is the work function of the metal surface.
This equation shows that the maximum kinetic energy depends only on the frequency of the incident light and the nature of the material (work function).
It does not depend on the intensity $(I)$ of the incident light.
Therefore,the graph between maximum kinetic energy $(E)$ and intensity $(I)$ should be a horizontal line parallel to the intensity axis.
Looking at the given graph,line $P$ represents a constant value of $E$ regardless of the intensity $I$.
Thus,line $P$ is the correct representation.

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Substituting the values: $E = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3315 \times 10^{-10}} = 6 \times 10^{-19} \ \text{J}$.
According to Einstein's photoelectric equation, $E = \phi_{0} + K_{\text{max}}$, where $\phi_{0}$ is the work function and $K_{\text{max}}$ is the kinetic energy of the ejected electron.
$\phi_{0} = E - K_{\text{max}} = 6 \times 10^{-19} - 3 \times 10^{-19} = 3 \times 10^{-19} \ \text{J}$.
The threshold wavelength $\lambda_{0}$ is given by $\lambda_{0} = \frac{hc}{\phi_{0}}$.
$\lambda_{0} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 10^{-19}} = 6.63 \times 10^{-7} \ \text{m} = 6630 \ \text{Å}$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \phi_{0}$,where $E$ is the incident energy and $\phi_{0}$ is the work function.
For the first radiation,$E_{1} = 2\phi_{0}$,so $K_{1} = \frac{1}{2}mv_{1}^{2} = 2\phi_{0} - \phi_{0} = \phi_{0}$.
For the second radiation,$E_{2} = 10\phi_{0}$,so $K_{2} = \frac{1}{2}mv_{2}^{2} = 10\phi_{0} - \phi_{0} = 9\phi_{0}$.
Taking the ratio of the two equations: $\frac{\frac{1}{2}mv_{1}^{2}}{\frac{1}{2}mv_{2}^{2}} = \frac{\phi_{0}}{9\phi_{0}}$.
This simplifies to $\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{9}$.
Taking the square root on both sides,we get $\frac{v_{1}}{v_{2}} = \frac{1}{3}$.

(C) The work function $\phi$ of a photosensitive surface is a characteristic property of the material itself and does not depend on the wavelength of the incident light.
Since the same photosensitive surface is used for both wavelengths,the work function remains constant.
Therefore,the ratio of the work functions for the two cases is $\phi_{1} : \phi_{2} = 1 : 1$.
However,if the question implies the ratio of the energy of the incident photons,it would be $\frac{E_{1}}{E_{2}} = \frac{hc/\lambda_{1}}{hc/\lambda_{2}} = \frac{\lambda_{2}}{\lambda_{1}} = \frac{600}{360} = \frac{5}{3}$.
Given the standard interpretation of such problems in physics textbooks,the ratio of the work functions for the same material is always $1:1$. Since $1:1$ is not an option,the question likely asks for the ratio of the energy of the incident photons,which is $5:3$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{\max} = eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.
For the first case: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$ --- $(i)$
For the second case: $e(\frac{V}{3}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$ --- (ii)
Dividing equation $(i)$ by equation (ii):
$\frac{eV}{eV/3} = \frac{\frac{hc}{\lambda} - \frac{hc}{\lambda_0}}{\frac{hc}{2\lambda} - \frac{hc}{\lambda_0}}$
$3 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$
$3(\frac{1}{2\lambda} - \frac{1}{\lambda_0}) = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{3}{2\lambda} - \frac{3}{\lambda_0} = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{3}{2\lambda} - \frac{1}{\lambda} = \frac{3}{\lambda_0} - \frac{1}{\lambda_0}$
$\frac{1}{2\lambda} = \frac{2}{\lambda_0}$
$\lambda_0 = 4\lambda$.

(B) The stopping potential $(V_0)$ is the minimum negative potential applied to the anode with respect to the cathode that stops even the most energetic photoelectrons from reaching the collector.
According to Einstein's photoelectric equation: $K_{max} = h\nu - \Phi_0$, where $K_{max} = eV_0$.
Thus, $eV_0 = h\nu - \Phi_0$, which implies $V_0 = \frac{h}{e}\nu - \frac{\Phi_0}{e}$.
From this equation, it is clear that the stopping potential $V_0$ is a linear function of the frequency of incident light ($\nu$).
Therefore, the stopping potential is directly proportional to the frequency of incident light (above the threshold frequency).

(D) The maximum kinetic energy of photoelectrons is given by Einstein's photoelectric equation: $KE_{\text{max}} = h(v - v_0)$ ... $(i)$
Where $h$ is Planck's constant,$v$ is the frequency of incident radiation,and $v_0$ is the threshold frequency.
From equation $(i)$,it is clear that the maximum kinetic energy $(KE_{\text{max}})$ depends only on the frequency of the incident light and the work function of the metal.
It is independent of the intensity of the incident radiation.
Therefore,if the intensity is reduced to one-fourth of its original value,the maximum kinetic energy of the emitted photoelectrons remains unchanged.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by:
$K = hv - W_0$ --- $(1)$
When the frequency of the radiation is doubled,the new frequency becomes $2v$. The new maximum kinetic energy $K'$ is:
$K' = h(2v) - W_0$
$K' = 2hv - W_0$ --- $(2)$
From equation $(1)$,we have $hv = K + W_0$. Substituting this into equation $(2)$:
$K' = 2(K + W_0) - W_0$
$K' = 2K + 2W_0 - W_0$
$K' = 2K + W_0$
Alternatively,using the difference method:
$K' - K = (2hv - W_0) - (hv - W_0) = hv$
$K' = K + hv$

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by:
$K_{max} = hf - \phi$
Since the stopping potential $(V_s)$ is related to the maximum kinetic energy by $K_{max} = eV_s$,we can write:
$eV_s = hf - \phi$
$V_s = (h/e)f - (\phi/e)$
This equation is of the form $y = mx + c$,which represents a straight line.
Here,the slope is $(h/e)$ (a positive constant) and the intercept on the $V_s$-axis is $(-\phi/e)$.
When $V_s = 0$,we have $hf = \phi$,which gives $f = \phi/h = f_0$ (the threshold frequency).
Thus,the graph of $V_s$ versus $f$ is a straight line starting from $f = f_0$ on the frequency axis.
Graph $(1)$ correctly represents this linear relationship.

(B) According to Einstein's photoelectric equation: $e V_0 = h f - \Phi$, where $\Phi = h f_0$ is the work function.
Rearranging the equation for stopping potential $V_0$: $V_0 = \frac{h}{e} f - \frac{h f_0}{e}$.
Since $h$, $e$, and $f_0$ are constants, $V_0$ is a linear function of the incident frequency $f$.
As the frequency $f$ of the incident radiation increases, the term $\frac{h}{e} f$ increases.
Therefore, the stopping potential $V_0$ increases.

(B) According to Einstein's photoelectric equation: $K_{max} = h\nu - W_0$,where $W_0$ is the work function.
Initially: $0.4 = h\nu - W_0 \implies h\nu = 0.4 + W_0$ ... $(i)$
When frequency is increased by $30 \%$,the new frequency $\nu' = 1.3\nu$. The new kinetic energy is $0.9 \ eV$.
So,$0.9 = 1.3h\nu - W_0$ ... (ii)
Substitute $(i)$ into (ii): $0.9 = 1.3(0.4 + W_0) - W_0$
$0.9 = 0.52 + 1.3W_0 - W_0$
$0.9 - 0.52 = 0.3W_0$
$0.38 = 0.3W_0$
$W_0 = \frac{0.38}{0.3} = 1.267 \ eV$.

(C) The energy of an electron in the $n^{\text{th}}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$.
For the ground state $(n=1)$,$E_1 = -13.6 \ eV$.
For the second orbit $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \ eV$.
The energy of the emitted photon is the difference between these two states: $E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$.
According to Einstein's photoelectric equation,$h\nu = \phi_0 + eV_s$,where $h\nu$ is the photon energy,$\phi_0$ is the work function,and $V_s$ is the stopping potential.
Given $\phi_0 = 4.2 \ eV$ and $h\nu = 10.2 \ eV$,we have $10.2 \ eV = 4.2 \ eV + eV_s$.
$eV_s = 10.2 \ eV - 4.2 \ eV = 6 \ eV$.
Therefore,the stopping potential $V_s = 6 \ V$.

(A) According to the photoelectric equation: $\frac{hc}{\lambda} = \phi + eV_s$.
Here,$V_s$ is the stopping potential and $\phi$ is the work function.
The kinetic energy of the emitted photoelectrons is given by $K_{max} = eV_s = \frac{1}{2}mv^2$.
If the velocity $v$ of the emitted photoelectrons is kept constant,the kinetic energy $K_{max}$ remains constant.
However,as the incident wavelength $\lambda$ is decreased,the energy of the incident photon $E = \frac{hc}{\lambda}$ increases.
Since $E = \phi + K_{max}$,and $K_{max}$ is constant while $E$ increases,the work function $\phi$ of the material would effectively need to change or the stopping potential $V_s$ must increase to balance the equation if we consider the energy conservation principle for a fixed material.
Actually,for a fixed material,if $\lambda$ decreases,$K_{max}$ increases. If the problem states the velocity is kept constant,it implies an external adjustment or a misunderstanding of the physical constraints; however,based on the standard photoelectric equation $\frac{hc}{\lambda} = \phi + eV_s$,if $\lambda$ decreases,the energy of the incident photon increases,which leads to an increase in the stopping potential $V_s$.

(D) According to Einstein's photoelectric equation,$K_{\max} = h\nu - \Phi$,where $h\nu$ is the energy of the incident photon and $\Phi$ is the work function of the metal.
$K_{\max}$ depends only on the frequency of the incident light and the nature of the metal surface.
Intensity of light is related to the number of photons incident per unit area per unit time,which affects the number of photoelectrons emitted (photoelectric current),but not their individual maximum kinetic energy.
Therefore,if the intensity of light is doubled,the maximum kinetic energy of the photoelectrons remains unchanged.