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

(D) The photoelectric effect occurs when the incident wavelength $\lambda$ is less than or equal to the threshold wavelength $\lambda_{\text{th}}$.
Given:
Threshold wavelength $\lambda_{\text{th}} = 3600 \mathring A$.
Incident wavelength $\lambda = 1100 \mathring A$.
The kinetic energy $K_{\text{max}}$ is given by Einstein's photoelectric equation:
$K_{\text{max}} = \frac{hc}{\lambda} - \frac{hc}{\lambda_{\text{th}}}$
Using $hc \approx 12400 \text{ eV} \cdot \mathring A$:
$K_{\text{max}} = 12400 \left( \frac{1}{1100} - \frac{1}{3600} \right) \text{ eV}$
$K_{\text{max}} = 12400 \left( \frac{3600 - 1100}{1100 \times 3600} \right) \text{ eV}$
$K_{\text{max}} = 12400 \left( \frac{2500}{3960000} \right) \text{ eV} \approx 7.83 \text{ eV}$.

(C) The work function $\Phi_0$ is given by the formula $\Phi_0 = \frac{hc}{\lambda_0}$,where $h$ is Planck's constant $(6.63 \times 10^{-34} \ J \cdot s)$,$c$ is the speed of light $(3 \times 10^8 \ m/s)$,and $\lambda_0$ is the threshold wavelength $(5000 \ Å = 5000 \times 10^{-10} \ m = 5 \times 10^{-7} \ m)$.
Substituting the values:
$\Phi_0 = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{5 \times 10^{-7}}$
$\Phi_0 = \frac{19.89 \times 10^{-26}}{5 \times 10^{-7}}$
$\Phi_0 = 3.978 \times 10^{-19} \ J \approx 4 \times 10^{-19} \ J$.
Thus,the correct option is $C$.

(B) Let the work function of the emitter plate be $\phi = h v_0$. The energy of the incident photon is $E = h v$.
Case $1$: Switch $S_1$ is closed and $S_2$ is open. The battery of $5 V$ is connected such that the collector $C$ is at a potential of $+5 V$ relative to $E$. The maximum kinetic energy of photoelectrons at the collector is $K_{max, C} = K_{max, E} + e V_{acc} = (E - \phi) + 5 eV = 1 eV$.
Thus,$E - \phi = -4 eV$ (This implies the light frequency is below the threshold,but the problem states photoelectrons strike,so we interpret the potential as retarding). Looking at the circuit,$S_1$ closed means $C$ is at $-5 V$ relative to $E$. So,$K_{max, C} = (E - \phi) - 5 eV = 1 eV \Rightarrow E - \phi = 6 eV$.
Case $2$: Switch $S_1$ is open and $S_2$ is closed. The battery of $5 V$ is connected such that $C$ is at $+5 V$ relative to $E$. The frequency is doubled,so the new energy is $2E$. The maximum kinetic energy at the collector is $K_{max, C} = (2E - \phi) + 5 eV = 20 eV \Rightarrow 2E - \phi = 15 eV$.
We have the system of equations:
$E - \phi = 6 eV$ ...$(i)$
$2E - \phi = 15 eV$ ...(ii)
Subtracting $(i)$ from (ii): $E = 9 eV$.
Substituting $E = 9 eV$ into $(i)$: $9 eV - \phi = 6 eV \Rightarrow \phi = 3 eV$.
The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \frac{hc}{\phi} = \frac{12400 eV Å}{3 eV} \approx 4133.3 Å$.

(D) Given,work function $W_0 = 2.35 \text{ eV}$. The electric component of the electromagnetic radiation is given by $E = a[1 + \cos(2 \pi f_1 t)] \cos(2 \pi f_2 t)$.
Using the trigonometric identity $\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$,we expand the expression:
$E = a \cos(2 \pi f_2 t) + a \cos(2 \pi f_1 t) \cos(2 \pi f_2 t)$
$E = a \cos(2 \pi f_2 t) + \frac{a}{2} \cos[2 \pi (f_1 + f_2) t] + \frac{a}{2} \cos[2 \pi (f_1 - f_2) t]$.
The frequencies present in the radiation are $f_2$,$(f_1 + f_2)$,and $(f_1 - f_2)$.
To obtain the maximum kinetic energy,we must use the photon with the maximum frequency,which is $f_{\max} = f_1 + f_2 = 3.6 \times 10^{15} + 1.2 \times 10^{15} = 4.8 \times 10^{15} \text{ Hz}$.
The energy of this photon is $E_{\text{photon}} = h f_{\max} = (6.6 \times 10^{-34} \text{ Js} \times 4.8 \times 10^{15} \text{ Hz}) / (1.6 \times 10^{-19} \text{ J/eV}) = 19.8 \text{ eV}$.
The maximum kinetic energy is $KE_{\max} = E_{\text{photon}} - W_0 = 19.8 \text{ eV} - 2.35 \text{ eV} = 17.45 \text{ eV}$.
Thus,the correct option is $D$.

(A) Given,wavelength of incident radiation,$\lambda = 400 \,nm = 4 \times 10^{-7} \,m$.
Electric field,$E = 2 \,N/C$ and distance,$s = 1 \,m$.
Energy of the incident photon,$E_{ph} = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} \approx 4.97 \times 10^{-19} \,J$.
Converting to $eV$,$E_{ph} = \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.1 \,eV$.
The electrons come to rest after traveling $1 \,m$ against the electric field. The work done by the electric field on the electron is $W = qEs = (1.6 \times 10^{-19} \,C) \times (2 \,N/C) \times (1 \,m) = 3.2 \times 10^{-19} \,J$.
Converting this to $eV$,$K_{max} = 2 \,eV$.
Using Einstein's photoelectric equation,$E_{ph} = W_0 + K_{max}$.
Therefore,the work function $W_0 = E_{ph} - K_{max} = 3.1 \,eV - 2 \,eV = 1.1 \,eV$.

(A) Given: Wavelength of light, $\lambda = 488 \, nm$ and stopping potential, $V_0 = 0.38 \, V$.
Energy of a photon is given by $E = \frac{hc}{\lambda}$.
Using $hc \approx 1240 \, eV \cdot nm$ (or $1242 \, eV \cdot nm$), we calculate $E = \frac{1240}{488} \approx 2.54 \, eV$.
Using $hc = 1242 \, eV \cdot nm$, $E = \frac{1242}{488} \approx 2.545 \, eV$.
Maximum kinetic energy of photoelectrons is $KE_{\max} = eV_0 = 0.38 \, eV$.
According to Einstein's photoelectric equation, $E = KE_{\max} + W_0$.
Therefore, the work function $W_0 = E - KE_{\max} = 2.54 - 0.38 = 2.16 \, eV$.
Thus, the work function of the cathode material is $2.16 \, eV$.

(A) The energy of a photon corresponding to a spectral line is given by $E = h\nu = \Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) eV$.
Threshold energy $\phi = E(H_\alpha) = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{5}{36} \right) eV$.
Energy of $H_\beta$ photon $E_\beta = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{3}{16} \right) eV$.
Energy of $H_\infty$ photon $E_\infty = 13.6 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = 13.6 \left( \frac{1}{4} \right) eV$.
Maximum kinetic energy $K_{max} = E - \phi$.
$K_\beta = 13.6 \left( \frac{3}{16} - \frac{5}{36} \right) = 13.6 \left( \frac{27 - 20}{144} \right) = 13.6 \left( \frac{7}{144} \right) eV$.
$K_\infty = 13.6 \left( \frac{1}{4} - \frac{5}{36} \right) = 13.6 \left( \frac{9 - 5}{36} \right) = 13.6 \left( \frac{4}{36} \right) = 13.6 \left( \frac{16}{144} \right) eV$.
Ratio $\frac{K_\beta}{K_\infty} = \frac{7/144}{16/144} = \frac{7}{16}$.

(B) The energy of a single photon is $E = \frac{hc}{\lambda}$.
The total number of photons emitted per second by the source of power $P$ is $N = \frac{P}{E} = \frac{P \lambda}{hc}$.
The number of electrons produced per second is $n = \frac{I}{e}$,where $I$ is the photoelectric current and $e$ is the charge of an electron.
The percentage of incident photons that produce current is given by $\eta = \frac{n}{N} \times 100$.
Substituting the values: $\eta = \frac{I/e}{P \lambda / hc} \times 100 = \frac{Ihc}{eP \lambda} \times 100$.

(D) According to Einstein's photoelectric equation,the stopping potential $V$ is given by:
$eV = h v - \Phi$
$V = (h/e) v - (\Phi/e)$
Here,$h$ is Planck's constant,$e$ is the charge of an electron,$v$ is the frequency of incident light,and $\Phi$ is the work function of the metal.
This equation represents a straight line $y = mx + c$,where the slope $m = h/e$ is constant for all metals,and the $y$-intercept is $-(\Phi/e)$.
The threshold frequency $v_0$ is given by $\Phi = h v_0$,so $v_0 = \Phi/h$.
Since the work function of $A$ is greater than that of $B$ $(\Phi_A > \Phi_B)$,the threshold frequency of $A$ must be greater than that of $B$ $(v_{0A} > v_{0B})$.
On a $V$ versus $v$ graph,the $x$-intercept represents the threshold frequency $v_0$.
Therefore,the line for metal $A$ must intersect the $v$-axis at a point further to the right than the line for metal $B$.
Looking at the provided options,the curve where $A$ has a larger threshold frequency than $B$ is the correct representation.

(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 $E = 5 eV$ is the energy of the incident photon and $\Phi = 4.36 eV$ is the work function of the metal.
$K_{max} = 5 eV - 4.36 eV = 0.64 eV$.
Converting this energy into Joules: $K_{max} = 0.64 \times 1.6 \times 10^{-19} J = 1.024 \times 10^{-19} J$.
The relationship between maximum kinetic energy and maximum momentum $p$ is $K_{max} = \frac{p^2}{2m}$,where $m$ is the mass of an electron $(9.1 \times 10^{-31} kg)$.
$p = \sqrt{2m K_{max}} = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.024 \times 10^{-19}}$.
$p = \sqrt{18.6368 \times 10^{-50}} \approx 4.31 \times 10^{-25} kgms^{-1}$.

(B) The energy of incident photons is $E = 4.8 eV$ and the work function of the sphere is $\phi = 2.8 eV$. Photoelectric emission stops when the potential of the sphere $V$ reaches a value such that the kinetic energy of the emitted electrons is zero. This occurs when $eV = E - \phi = 4.8 eV - 2.8 eV = 2.0 eV$,so $V = 2.0 V$.
The capacitance of a sphere of radius $R = 9 mm = 9 \times 10^{-3} m$ is $C = 4\pi\epsilon_0 R = \frac{R}{k} = \frac{9 \times 10^{-3}}{9 \times 10^9} = 10^{-12} F$.
The charge $Q$ required to reach potential $V$ is $Q = CV = 10^{-12} F \times 2.0 V = 2 \times 10^{-12} C$.
The number of electrons emitted is $n = \frac{Q}{e} = \frac{2 \times 10^{-12}}{1.6 \times 10^{-19}} = 1.25 \times 10^7$.
Since the rate of emission is $10^5$ electrons per second,the time taken is $t = \frac{n}{\text{rate}} = \frac{1.25 \times 10^7}{10^5} = 125 s$.

(B) According to Einstein's photoelectric equation: $K_{max} = h\nu - \phi_0$,where $\phi_0 = h\nu_0$.
For frequency $\nu_1 = 2\nu_0$,the maximum kinetic energy is $K_1 = h(2\nu_0) - h\nu_0 = h\nu_0$.
Given $v_1 = 2 \times 10^6 \ m/s$,so $K_1 = \frac{1}{2}mv_1^2 = h\nu_0$.
For frequency $\nu_2 = 3\nu_0$,the maximum kinetic energy is $K_2 = h(3\nu_0) - h\nu_0 = 2h\nu_0$.
Since $K_2 = 2K_1$,we have $\frac{1}{2}mv_2^2 = 2(\frac{1}{2}mv_1^2)$.
Thus,$v_2^2 = 2v_1^2$,which implies $v_2 = \sqrt{2}v_1$.
Substituting $v_1 = 2 \times 10^6 \ m/s$,we get $v_2 = \sqrt{2} \times 2 \times 10^6 = 2\sqrt{2} \times 10^6 \ m/s$.

(B) The energy of a photon emitted in the hydrogen spectrum is given by the Rydberg formula: $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}$.
For the Balmer series, the transition ends at $n_1 = 2$.
The maximum energy photon corresponds to the transition from $n_2 = \infty$ to $n_1 = 2$.
Substituting these values: $E_{\text{max}} = 13.6 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \text{ eV} = 13.6 \times \frac{1}{4} \text{ eV} = 3.4 \text{ eV}$.
For photoemission to occur, the energy of the incident photon must be greater than or equal to the work function $(\phi)$ of the metal. Thus, the maximum possible work function for which photoelectrons can be ejected is equal to the maximum energy of the incident photon, which is $3.4 \text{ eV}$.

(B) The energy of the photon released during the transition from the second excited state $(n_2 = 3)$ to the ground state $(n_1 = 1)$ is given by:
$E = 13.6 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \ eV$
$E = 13.6 \left[ \frac{1}{1^2} - \frac{1}{3^2} \right] = 13.6 \left[ 1 - \frac{1}{9} \right] = 13.6 \times \frac{8}{9} \approx 12.09 \ eV \approx 12.1 \ eV$.
Using Einstein's photoelectric equation,the maximum kinetic energy $(K_{\max})$ of the emitted electron is:
$K_{\max} = E - W = 12.1 \ eV - 3.1 \ eV = 9.0 \ eV$.
The de Broglie wavelength $(\lambda)$ is given by $\lambda = \frac{h}{\sqrt{2mK_{\max}}}$.
Using the formula $\lambda \approx \frac{12.27}{\sqrt{K_{\max}}} \ \text{Å}$ (where $K_{\max}$ is in $eV$):
$\lambda = \frac{12.27}{\sqrt{9}} = \frac{12.27}{3} = 4.09 \ \text{Å}$.
Thus,the wavelength is nearly $4 \ \text{Å}$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy of 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.
Since $K_{max} = e V_s$,where $V_s$ is the stopping potential,we have $e V_s = E - \phi$.
For the first case: $e(1.7 \ V) = 3.1 \ eV - \phi$,which implies $\phi = 3.1 \ eV - 1.7 \ eV = 1.4 \ eV$.
For the second case: $e V_s' = 2.5 \ eV - \phi$.
Substituting the value of $\phi$: $e V_s' = 2.5 \ eV - 1.4 \ eV = 1.1 \ eV$.
Therefore,the stopping potential $V_s' = 1.1 \ V$.

(A) The maximum kinetic energy $(K_{max})$ of the emitted photoelectrons is given by the formula: $K_{max} = \frac{1}{2} m v_{max}^2$.
Given $m = 9 \times 10^{-31} \text{ kg}$ and $v_{max} = 8 \times 10^5 \text{ m/s}$.
$K_{max} = \frac{1}{2} \times (9 \times 10^{-31}) \times (8 \times 10^5)^2$.
$K_{max} = 0.5 \times 9 \times 10^{-31} \times 64 \times 10^{10} = 288 \times 10^{-21} \text{ J}$.
To convert this energy into electron-volts (eV),divide by the charge of an electron $(e = 1.6 \times 10^{-19} \text{ C})$:
$K_{max} \text{ (in eV)} = \frac{288 \times 10^{-21}}{1.6 \times 10^{-19}} = 180 \times 10^{-2} = 1.8 \text{ eV}$.
The stopping potential $(V_s)$ is related to the maximum kinetic energy by the equation: $K_{max} = e V_s$.
Therefore,$V_s = \frac{K_{max}}{e} = 1.8 \text{ V}$.

(C) According to Einstein's photoelectric equation, the maximum kinetic energy of emitted 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.
Since $K_{max} = eV_s$, where $e$ is the charge of an electron and $V_s$ is the stopping potential, we can write:
$eV_s = h\nu - \phi_0$
$V_s = (\frac{h}{e})\nu - \frac{\phi_0}{e}$
This equation is in the form of a straight line $y = mx + c$, where $y = V_s$, $x = \nu$, and the slope $m = \frac{h}{e}$.
Given $h = 6.6 \times 10^{-34} \text{ Js}$ and $e = 1.6 \times 10^{-19} \text{ C}$.
Slope $m = \frac{6.6 \times 10^{-34}}{1.6 \times 10^{-19}} = 4.125 \times 10^{-15} \text{ JsC}^{-1}$.
Thus, the correct option is $C$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = h\nu - \phi_0$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\phi_0$ is the work function.
Since $K_{max} = eV_0$,where $V_0$ is the cut-off (stopping) potential and $e$ is the charge of an electron,we have $eV_0 = h\nu - \phi_0$.
Rearranging this,we get $V_0 = (\frac{h}{e})\nu - \frac{\phi_0}{e}$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m$ of the graph between cut-off voltage $V_0$ and frequency $\nu$ is $\frac{h}{e}$.
Substituting the values: $\text{Slope} = \frac{6.63 \times 10^{-34} \ J \cdot s}{1.6 \times 10^{-19} \ C} = 4.14 \times 10^{-15} \ V \cdot s$.

(C) The work function of the metal is $\phi_0 = 1.1 \ eV$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = E - \phi_0$,where $E$ is the energy of the incident photon.
Since $K_{max} = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2(E - \phi_0)}{m}}$.
For the first beam with energy $E_1 = 1.5 \ eV$,the maximum velocity is $v_1 = \sqrt{\frac{2(1.5 - 1.1)}{m}} = \sqrt{\frac{2(0.4)}{m}}$.
For the second beam with energy $E_2 = 2 \ eV$,the maximum velocity is $v_2 = \sqrt{\frac{2(2 - 1.1)}{m}} = \sqrt{\frac{2(0.9)}{m}}$.
The ratio of the maximum velocities is $\frac{v_1}{v_2} = \sqrt{\frac{0.4}{0.9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.
Thus,the ratio is $2:3$.

(A) Given:
Work function,$W = 2.2 \,eV$
Wavelength of incident radiation,$\lambda = 400 \,nm$
The energy of the incident photon is given by:
$E = \frac{1240 \,eV \cdot nm}{\lambda (nm)} = \frac{1240}{400} = 3.1 \,eV$
According to Einstein's photoelectric equation:
$E = W + K_{max}$
where $K_{max} = e V_s$ ($V_s$ is the stopping potential).
$e V_s = E - W$
$e V_s = 3.1 \,eV - 2.2 \,eV = 0.9 \,eV$
Therefore,the stopping potential $V_s = 0.9 \,V$.

(C) 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.
$1$. Statement $I$ is incorrect because $K_{\max}$ depends directly on the frequency $\nu$.
$2$. Statement $II$ is incorrect because the photoelectric effect depends on the frequency of light being greater than the threshold frequency $(\nu_0)$,regardless of intensity.
$3$. Statement $III$ is correct because $K_{\max}$ depends only on the frequency of incident light,not on its intensity.
$4$. Statement $IV$ is correct because as frequency $\nu$ increases,$K_{\max} = h\nu - \phi$ also increases.
Thus,statements $III$ and $IV$ are correct.

(C) From Einstein's photoelectric equation,the stopping potential $V_s$ is related to the frequency $\nu$ as:
$e V_s = h \nu - \phi$
Rearranging for $V_s$,we get:
$V_s = \left( \frac{h}{e} \right) \nu - \frac{\phi}{e}$
This is a linear equation of the form $y = mx + c$,where the slope $m = \frac{h}{e}$.
Given that the slope is $4 \times 10^{-15} \ V \ s$,we have:
$\frac{h}{e} = 4 \times 10^{-15} \ V \ s$
$h = (4 \times 10^{-15} \ V \ s) \times (1.6 \times 10^{-19} \ C)$
$h = 6.4 \times 10^{-34} \ J \ s$
Thus,the value of Planck's constant is $6.4 \times 10^{-34} \ J \ s$.

(D) According to Einstein's photoelectric equation, the maximum kinetic energy $(K.E._{max})$ of the emitted electrons is given by:
$K.E._{max} = E - W$
where $E$ is the energy of the incident photon and $W$ is the work function of the metal.
The energy of the incident photon is:
$E = \frac{hc}{\lambda} = \frac{1240 \, eV \cdot nm}{248 \, nm} = 5.0 \, eV$
The maximum kinetic energy is related to the stopping potential $(V_s)$ by:
$K.E._{max} = e V_s = 2.8 \, eV$
Substituting these values into the photoelectric equation:
$2.8 \, eV = 5.0 \, eV - W$
$W = 5.0 \, eV - 2.8 \, eV$
$W = 2.2 \, eV$
Thus, the work function of the metal is $2.2 \, eV$.

(A) The photoelectric equation is given by $eV_S = \frac{hc}{\lambda} - \Phi$, where $\Phi = \frac{hc}{\lambda_0}$ is the work function of the metal.
Thus, the stopping potential is $V_S = \frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$.
For the first wavelength $\lambda_1 = 200 \, nm$, the stopping potential is $V_{S1} = \frac{1240}{e} \left( \frac{1}{200} - \frac{1}{\lambda_0} \right)$.
For the second wavelength $\lambda_2 = 400 \, nm$, the stopping potential is $V_{S2} = \frac{1240}{e} \left( \frac{1}{400} - \frac{1}{\lambda_0} \right)$.
The decrease in stopping potential is $\Delta V_S = V_{S1} - V_{S2}$.
$\Delta V_S = \frac{1240}{e} \left( \frac{1}{200} - \frac{1}{\lambda_0} - \left( \frac{1}{400} - \frac{1}{\lambda_0} \right) \right)$.
$\Delta V_S = \frac{1240}{e} \left( \frac{1}{200} - \frac{1}{400} \right) = \frac{1240}{e} \left( \frac{2-1}{400} \right) = \frac{1240}{400} = 3.1 \, V$.

(A) The maximum kinetic energy $(K.E.)_{\max}$ of the emitted electron is given by the formula:
$(K.E.)_{\max} = \frac{1}{2} m v_{\max}^2$
Given $m = 9 \times 10^{-31} \ kg$ and $v_{\max} = 1.6 \times 10^6 \ m/s$,we have:
$(K.E.)_{\max} = \frac{1}{2} \times (9 \times 10^{-31}) \times (1.6 \times 10^6)^2$
$(K.E.)_{\max} = 0.5 \times 9 \times 10^{-31} \times 2.56 \times 10^{12}$
$(K.E.)_{\max} = 11.52 \times 10^{-19} \ J$
The stopping potential $V_s$ is related to the maximum kinetic energy by the equation:
$e V_s = (K.E.)_{\max}$
$V_s = \frac{(K.E.)_{\max}}{e} = \frac{11.52 \times 10^{-19} \ J}{1.6 \times 10^{-19} \ C}$
$V_s = 7.2 \ V$
Therefore,the correct option is $A$.

(A) According to Einstein's photoelectric equation,$h \nu - W = K_{\text{max}}$,where $h$ is Planck's constant,$\nu$ is the frequency of incident radiation,$W$ is the work function,and $K_{\text{max}}$ is the maximum kinetic energy of the emitted photoelectrons.
Since kinetic energy cannot be negative,$K_{\text{max}} \geq 0$.
Therefore,$h \nu - W \geq 0$,which implies $h \nu \geq W$.
Given that the work function $W = h \nu_0$,the condition becomes $h \nu \geq h \nu_0$.
Dividing both sides by $h$,we get $\nu \geq \nu_0$.
Thus,the photoelectric effect will take place only if the frequency of incident light is greater than or equal to the threshold frequency $\nu_0$.

(D) Given that,work function $\phi_0 = 1.5 \ eV$. The stopping potential $V_s$ is given by Einstein's photoelectric equation: $eV_s = \frac{hc}{\lambda} - \phi_0$. Since $V_s \propto \frac{1}{\lambda}$,smaller wavelength corresponds to a larger magnitude of stopping potential (more negative).
Comparing wavelengths: $\lambda_A = 200 \ nm < \lambda_B = 400 \ nm < \lambda_C = 600 \ nm$. Thus,the magnitude of stopping potentials follows $|V_A| > |V_B| > |V_C|$. From the graph,the stopping potentials are $V_1, V_2, V_3, V_4$ where $|V_1| > |V_2| > |V_3| > |V_4|$. Therefore,$A \rightarrow V_2$,$B \rightarrow V_3$,and $C \rightarrow V_4$.
Saturation photocurrent is directly proportional to the intensity of incident light. Given intensities: $I_A = 1.8 \ W/m^2$,$I_B = 1 \ W/m^2$,$I_C = 0.5 \ W/m^2$. Thus,the saturation current order is $I_A > I_B > I_C$.
Matching these with the curves:
Curve $III$ has stopping potential $V_2$ and high saturation current,corresponding to light $A$.
Curve $II$ has stopping potential $V_3$ and medium saturation current,corresponding to light $B$.
Curve $IV$ has stopping potential $V_4$ and low saturation current,corresponding to light $C$.
Thus,the correct correspondence is $A \rightarrow III, B \rightarrow II, C \rightarrow IV$. Option $(d)$ is correct.

(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 $\lambda_1 = 600 \ nm$,$K_1 = \frac{hc}{\lambda_1} - \phi$ --- $(i)$
For $\lambda_2 = 400 \ nm$,$K_2 = 2K_1 = \frac{hc}{\lambda_2} - \phi$ --- (ii)
From $(i)$,$K_1 = \frac{hc}{\lambda_1} - \phi$. Substituting this into (ii):
$2 \left( \frac{hc}{\lambda_1} - \phi \right) = \frac{hc}{\lambda_2} - \phi$
$\frac{2hc}{\lambda_1} - 2\phi = \frac{hc}{\lambda_2} - \phi$
$\phi = hc \left( \frac{2}{\lambda_1} - \frac{1}{\lambda_2} \right)$
Substituting the values: $h = 6.63 \times 10^{-34} \ J-s$,$c = 3 \times 10^8 \ m/s$,$\lambda_1 = 600 \times 10^{-9} \ m$,$\lambda_2 = 400 \times 10^{-9} \ m$:
$\phi = (6.63 \times 10^{-34} \times 3 \times 10^8) \left( \frac{2}{600 \times 10^{-9}} - \frac{1}{400 \times 10^{-9}} \right)$
$\phi = (19.89 \times 10^{-26}) \times 10^9 \left( \frac{1}{300} - \frac{1}{400} \right)$
$\phi = 19.89 \times 10^{-17} \left( \frac{4-3}{1200} \right) = \frac{19.89 \times 10^{-17}}{1200} \ J$
$\phi = \frac{19.89 \times 10^{-17}}{1200 \times 1.6 \times 10^{-19}} \ eV \approx 1.036 \ eV \approx 1.02 \ eV$.

(A) Let $\phi_0$ be the work function of the metal.
According to Einstein's photoelectric equation, the maximum kinetic energy $K_{\max}$ is given by:
$K_{\max} = \frac{1}{2} m v^2 = \frac{hc}{\lambda} - \phi_0$ --- $(i)$
Given that the initial energy of the photon is $E = \frac{hc}{\lambda} = 2.4 \text{ eV}$.
When the wavelength is decreased by $50 \%$, the new wavelength is $\lambda' = \frac{\lambda}{2}$.
The new energy of the photon is $E' = \frac{hc}{\lambda'} = \frac{hc}{\lambda / 2} = 2 \left( \frac{hc}{\lambda} \right) = 2 \times 2.4 \text{ eV} = 4.8 \text{ eV}$.
The new maximum velocity is $v' = 3v$, so the new maximum kinetic energy is $K'_{\max} = \frac{1}{2} m (3v)^2 = 9 \left( \frac{1}{2} m v^2 \right) = 9 K_{\max}$.
Using the photoelectric equation for the second case:
$9 K_{\max} = E' - \phi_0 = 4.8 - \phi_0$ --- (ii)
From equation $(i)$, $K_{\max} = 2.4 - \phi_0$.
Substituting this into equation (ii):
$9(2.4 - \phi_0) = 4.8 - \phi_0$
$21.6 - 9\phi_0 = 4.8 - \phi_0$
$8\phi_0 = 21.6 - 4.8 = 16.8$
$\phi_0 = \frac{16.8}{8} = 2.1 \text{ eV}$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of emitted electrons is given by $K_{\max} = \frac{1}{2} m v_{\max}^2 = E - \phi_0$,where $E$ is the incident photon energy and $\phi_0$ is the work function.
For the first case,$E_1 = 4 \text{ eV}$ and $\phi_0 = 2 \text{ eV}$:
$\frac{1}{2} m v_1^2 = 4 - 2 = 2 \text{ eV} \quad \dots(i)$
For the second case,$E_2 = 2.5 \text{ eV}$ and $\phi_0 = 2 \text{ eV}$:
$\frac{1}{2} m v_2^2 = 2.5 - 2 = 0.5 \text{ eV} \quad \dots(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{2}{0.5}$
$\frac{v_1^2}{v_2^2} = 4$
Taking the square root on both sides:
$\frac{v_1}{v_2} = \sqrt{4} = 2$

(B) According to Einstein's photoelectric equation,$E_K = E - \phi_0$,where $E_K$ is the kinetic energy,$E$ is the energy of the incident photon,and $\phi_0$ is the work function of the metal surface.
From this,we have $E = E_K + \phi_0$ ... $(i)$.
When the energy of the incident photon is increased by $20 \%$,the new energy $E'$ becomes $E' = E + 0.2E = 1.2E$.
The new kinetic energy $E_K'$ is given by $E_K' = E' - \phi_0$,so $E' = E_K' + \phi_0$ ... (ii).
Substituting $E' = 1.2E$ into equation (ii),we get $1.2E = E_K' + \phi_0$.
From equation $(i)$,$E = 0.5 + \phi_0$. Substituting this into the modified equation (ii):
$1.2(0.5 + \phi_0) = 0.8 + \phi_0$
$0.6 + 1.2\phi_0 = 0.8 + \phi_0$
$1.2\phi_0 - \phi_0 = 0.8 - 0.6$
$0.2\phi_0 = 0.2$
$\phi_0 = 1.0 \ eV$.

(B) Given, work function $\phi = 2.4 \text{ eV}$.
Energy required to ionize a hydrogen atom in the ground state is $E_i = 13.6 \text{ eV}$.
For the hydrogen atom to be ionized by the photoelectron, the kinetic energy $(KE)$ of the photoelectron must be at least equal to the ionization energy of the hydrogen atom.
Thus, $KE_{max} = 13.6 \text{ eV}$.
According to the photoelectric equation, $E = KE_{max} + \phi$, where $E = \frac{hc}{\lambda}$.
Substituting the values: $\frac{1240}{\lambda} = 13.6 + 2.4$.
$\frac{1240}{\lambda} = 16$.
$\lambda = \frac{1240}{16} = 77.5 \text{ nm}$.

(B) The energy equation in the photoelectric effect is given by Einstein's photoelectric equation:
$e V_0 = h \nu - \phi_0 \Rightarrow V_0 = \frac{h \nu}{e} - \frac{\phi_0}{e}$ $\ldots$ $(i)$
Where $V_0$ is the stopping potential,$h$ is Planck's constant,$\nu$ is the frequency,and $\phi_0$ is the work function.
When the frequency is doubled,the new frequency is $\nu' = 2\nu$.
The new stopping potential $V_0'$ is given by:
$V_0' = \frac{h(2\nu)}{e} - \frac{\phi_0}{e} = \frac{2h\nu}{e} - \frac{\phi_0}{e}$
We can rewrite this as:
$V_0' = 2\left(\frac{h\nu}{e} - \frac{\phi_0}{e}\right) + \frac{\phi_0}{e}$
Substituting equation $(i)$ into this expression:
$V_0' = 2V_0 + \frac{\phi_0}{e}$
Since $\frac{\phi_0}{e} > 0$,it follows that $V_0' > 2V_0$.
Therefore,the stopping potential will become more than double.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max }$ of emitted photoelectrons is given by: $(KE)_{\max } = h\nu - \phi_0$,where $h$ is Planck's constant,$\nu$ is the frequency of incident radiation,and $\phi_0$ is the work function of the metal.
Comparing this with the equation of a straight line $y = mx + c$,where $y = (KE)_{\max }$ and $x = \nu$,we get the slope $m = h$.
Since $h$ (Planck's constant) is a universal constant,the slope of the $(KE)_{\max }$ versus $\nu$ graph is the same for all metals and is independent of the intensity of the incident radiation.

(B) The work function of the metal surface is $\phi_0 = 2.5 \,eV$.
The wavelength of the incident light is $\lambda = 310 \,nm$.
Given the constant $hc = 1240 \,eV-nm$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by:
$K_{max} = \frac{hc}{\lambda} - \phi_0$
Since the stopping potential $V_0$ is related to the maximum kinetic energy by $K_{max} = eV_0$,we have:
$eV_0 = \frac{1240 \,eV-nm}{310 \,nm} - 2.5 \,eV$
$eV_0 = 4 \,eV - 2.5 \,eV$
$eV_0 = 1.5 \,eV$
Therefore,the stopping potential $V_0 = 1.5 \,V$.

(A) Given: Power of source $P = 1 \,W$, distance $d = 1 \,m$, radius of exposed area $r = 1 \,Å = 10^{-10} \,m$, work function $\phi = 5 \,eV = 5 \times 1.6 \times 10^{-19} \,J$.
Intensity $I$ at distance $d$ is $I = \frac{P}{4 \pi d^2} = \frac{1}{4 \pi (1)^2} = \frac{1}{4 \pi} \,W/m^2$.
Power absorbed by the circular area $A = \pi r^2$ is $P_{abs} = I \times A = \frac{1}{4 \pi} \times \pi (10^{-10})^2 = \frac{10^{-20}}{4} \,W$.
The time $t$ required to accumulate energy equal to the work function is $t = \frac{\phi}{P_{abs}}$.
Substituting the values: $t = \frac{5 \times 1.6 \times 10^{-19}}{10^{-20} / 4} = \frac{8 \times 10^{-19}}{0.25 \times 10^{-20}} = 320 \,s$.

(D) According to Einstein's photoelectric equation,$h v = \phi + K.E_{max}$,where $K.E_{max} = e V_s$.
For frequency $v_1$: $h v_1 = \phi + e V_1$ --- $(1)$
For frequency $2 v_1$: $h(2 v_1) = \phi + e(3 V_1) = \phi + 3 e V_1$ --- $(2)$
From $(1)$,$h v_1 = \phi + e V_1$. Substituting this into $(2)$:
$2(\phi + e V_1) = \phi + 3 e V_1$
$2 \phi + 2 e V_1 = \phi + 3 e V_1$
$\phi = e V_1$
Now,substitute $\phi = e V_1$ back into $(1)$:
$h v_1 = e V_1 + e V_1 = 2 e V_1$
For frequency $4 v_1$,let the stopping potential be $V_s = n V_1$:
$h(4 v_1) = \phi + e(n V_1)$
$4(h v_1) = e V_1 + n e V_1$
$4(2 e V_1) = e V_1(1 + n)$
$8 e V_1 = e V_1(1 + n)$
$8 = 1 + n$
$n = 7$

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Given $\lambda = 221 \,nm = 221 \times 10^{-9} \,m$, $h = 6.63 \times 10^{-34} \,J \cdot s$, and $c = 3 \times 10^8 \,m/s$.
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{221 \times 10^{-9}} \,J = \frac{19.89 \times 10^{-26}}{221 \times 10^{-9}} \,J = 0.09 \times 10^{-17} \,J = 9 \times 10^{-19} \,J$.
Converting this energy into electron-volts $(eV)$:
$E = \frac{9 \times 10^{-19} \,J}{1.6 \times 10^{-19} \,J/eV} = 5.625 \,eV \approx 5.63 \,eV$.
The work function of the lead ball is $\phi = 4.14 \,eV$.
The maximum kinetic energy of the emitted photoelectrons is $K_{max} = E - \phi = 5.63 \,eV - 4.14 \,eV = 1.49 \,eV$.
As the ball loses electrons, it becomes positively charged and its potential increases until the photoelectrons can no longer escape. The maximum potential $V$ attained is given by $eV = K_{max}$.
Thus, $V = 1.49 \,V$.

(A) According to Einstein's photoelectric equation, the energy of the incident photon $(E)$ is equal to the sum of the work function $(\Phi)$ and the maximum kinetic energy $(K_{max})$ of the emitted electron.
$E = \Phi + K_{max}$
Given:
Energy of incident photon $(E)$ = $4 eV$
Maximum kinetic energy $(K_{max})$ = $1.1 eV$
Rearranging the formula to find the work function:
$\Phi = E - K_{max}$
$\Phi = 4 eV - 1.1 eV$
$\Phi = 2.9 eV$
Therefore, the work function of the metal is $2.9 eV$.

(D) The energy of an incident photon is given by $E = hf$.
Given $h = 6.63 \times 10^{-34} \,J-s$ and $f = 4 \times 10^{14} \,Hz$.
$E = 6.63 \times 10^{-34} \times 4 \times 10^{14} = 26.52 \times 10^{-20} \,J$.
To convert this energy into electron-volts $(eV)$, divide by $1.6 \times 10^{-19} \,J/eV$:
$E = \frac{26.52 \times 10^{-20}}{1.6 \times 10^{-19}} \,eV = 1.6575 \,eV$.
The work function of the metal is $\Phi = 2.14 \,eV$.
Since the energy of the incident photon $(1.6575 \,eV)$ is less than the work function $(2.14 \,eV)$, the incident light does not have sufficient energy to eject electrons from the metal surface.
Therefore, no photoemission occurs, and the maximum kinetic energy is $0 \,eV$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE_{max})$ is given by:
$KE_{max} = \frac{hc}{\lambda} - \phi_0$
Given:
Work function $\phi_0 = 2 \ \text{eV} = 2 \times 1.6 \times 10^{-19} \ \text{J} = 3.2 \times 10^{-19} \ \text{J}$
Wavelength $\lambda = 3000 \ \text{Å} = 3000 \times 10^{-10} \ \text{m} = 3 \times 10^{-7} \ \text{m}$
Energy of incident photon $E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{3 \times 10^{-7}} \ \text{J} = 6.6 \times 10^{-19} \ \text{J}$
Now,calculate $KE_{max}$:
$KE_{max} = 6.6 \times 10^{-19} \ \text{J} - 3.2 \times 10^{-19} \ \text{J} = 3.4 \times 10^{-19} \ \text{J}$

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $E$ of an emitted electron is given by $E = \frac{hc}{\lambda} - W$,where $W$ is the work function of the metal.
For the first case: $E_1 = \frac{hc}{\lambda_1} - W$ --- $(i)$
For the second case: $E_2 = \frac{hc}{\lambda_2} - W$ --- (ii)
Subtracting equation (ii) from equation $(i)$:
$E_1 - E_2 = \left(\frac{hc}{\lambda_1} - W\right) - \left(\frac{hc}{\lambda_2} - W\right)$
$E_1 - E_2 = hc \left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)$
$E_1 - E_2 = hc \left(\frac{\lambda_2 - \lambda_1}{\lambda_1 \lambda_2}\right)$
Rearranging to solve for $hc$:
$hc = \frac{(E_1 - E_2) \lambda_1 \lambda_2}{\lambda_2 - \lambda_1}$

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of an emitted photoelectron is given by $K_{max} = E - \phi_0$,where $E$ is the energy of the incident photon and $\phi_0$ is the work function of the metal.
For the first photon with energy $E_1 = 2.5 eV$:
$\frac{1}{2} m v_1^2 = E_1 - \phi_0 = 2.5 eV - 1.5 eV = 1.0 eV$ $(i)$
For the second photon with energy $E_2 = 3.5 eV$:
$\frac{1}{2} m v_2^2 = E_2 - \phi_0 = 3.5 eV - 1.5 eV = 2.0 eV$ (ii)
Dividing equation $(i)$ by equation (ii),we get:
$\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{1.0 eV}{2.0 eV}$
$\frac{v_1^2}{v_2^2} = \frac{1}{2}$
Taking the square root on both sides:
$\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}$
Thus,the ratio of the maximum velocities is $1 : \sqrt{2}$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = h v - h v_0$,where $h$ is Planck's constant and $v_0$ is the threshold frequency.
For frequency $v_1$,$K_1 = h(v_1 - v_0)$.
For frequency $v_2$,$K_2 = h(v_2 - v_0)$.
Given the ratio of maximum kinetic energies is $K_1 : K_2 = 1 : n$,we have $\frac{K_1}{K_2} = \frac{1}{n}$.
Substituting the expressions,we get $\frac{h(v_1 - v_0)}{h(v_2 - v_0)} = \frac{1}{n}$.
$\frac{v_1 - v_0}{v_2 - v_0} = \frac{1}{n}$.
Cross-multiplying gives $n(v_1 - v_0) = v_2 - v_0$.
$n v_1 - n v_0 = v_2 - v_0$.
$n v_1 - v_2 = n v_0 - v_0$.
$n v_1 - v_2 = v_0(n - 1)$.
Therefore,the threshold frequency is $v_0 = \frac{n v_1 - v_2}{n - 1}$.

(A) According to Einstein's photoelectric equation,the kinetic energy $K$ of the ejected photoelectron is given by $K = E - W_0$.
Since $K = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2(E-W_0)}{m}}$.
When a charged particle of mass $m$ and charge $e$ moves in a uniform magnetic field $B$ perpendicular to its velocity,it experiences a magnetic Lorentz force which provides the necessary centripetal force for circular motion:
$evB = \frac{mv^2}{r}$
Solving for the radius $r$,we get $r = \frac{mv}{eB}$.
Substituting the expression for $v$ into the equation for $r$:
$r = \frac{m}{eB} \sqrt{\frac{2(E-W_0)}{m}} = \frac{\sqrt{m^2 \cdot \frac{2(E-W_0)}{m}}}{eB} = \frac{\sqrt{2m(E-W_0)}}{eB}$.