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

(B) The work function $\phi_{0}$ is given by the formula $\phi_{0} = h \nu_{0}$,where $h$ is Planck's constant $(6.626 \times 10^{-34} \ J \cdot s)$ and $\nu_{0}$ is the threshold frequency.
To convert the energy from Joules to electron-volts $(eV)$,we divide by the charge of an electron $(e = 1.6 \times 10^{-19} \ C)$.
$\phi_{0} = \frac{h \nu_{0}}{e} = \frac{6.626 \times 10^{-34} \times 5.16 \times 10^{14}}{1.6 \times 10^{-19}} \ eV$.
$\phi_{0} \approx \frac{34.19 \times 10^{-20}}{1.6 \times 10^{-19}} \ eV$.
$\phi_{0} \approx 2.137 \ eV$.
Rounding to two decimal places,we get $\phi_{0} \approx 2.14 \ eV$.

(A) The relationship between the maximum kinetic energy $(K_{\text{max}})$ of photoelectrons and the stopping potential (cut-off voltage,$V_0$) is given by the equation:
$K_{\text{max}} = e V_0$
Given that the cut-off voltage $V_0 = 1.5 \ V$,we substitute this value into the equation:
$K_{\text{max}} = e \times 1.5 \ V$
$K_{\text{max}} = 1.5 \ eV$
Therefore,the maximum kinetic energy of the emitted photoelectrons is $1.5 \ eV$.

(C) According to Einstein's photoelectric equation, $K_{max} = h\nu - \phi_0$, where $K_{max}$ is the maximum kinetic energy of emitted photoelectrons, $h$ is Planck's constant, $\nu$ is the frequency of incident light, and $\phi_0$ is the work function of the metal.
Since $K_{max} = eV_s$ (where $e$ is the charge of an electron and $V_s$ is the stopping potential), we have $eV_s = h\nu - \phi_0$.
Rearranging for the stopping potential, $V_s = \frac{h}{e}\nu - \frac{\phi_0}{e}$.
As the frequency $\nu$ of the incident light increases, the term $\frac{h}{e}\nu$ increases, which leads to an increase in the stopping potential $V_s$.
The photoelectric current depends on the intensity of the incident light, not its frequency, provided the frequency is above the threshold frequency.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = h\nu - \phi_{0}$.
Since the stopping potential $V_{0}$ is related to the maximum kinetic energy by $K_{max} = eV_{0}$,we can write $eV_{0} = h\nu - \phi_{0}$.
Dividing by the elementary charge $e$,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 $y = V_{0}$,$x = \nu$,and $c = -\frac{\phi_{0}}{e}$,the slope $m$ is equal to $\frac{h}{e}$.

(C) Given,energy of incident light is $E$ and work function is $\phi_0 = \frac{E}{3}$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:
$K_{\max} = E - \phi_0$
$K_{\max} = E - \frac{E}{3} = \frac{2E}{3}$.
Since photoelectrons are emitted with a range of energies from $0$ to $K_{\max}$ due to collisions within the metal,the kinetic energy $K$ of the photoelectrons lies in the range $0 \leq K \leq \frac{2E}{3}$.

(C) Given,work function $\phi_0 = 2.4 \ eV$.
Photoemission occurs only if the incident photon energy $E \ge \phi_0$,which implies the incident wavelength $\lambda \le \lambda_0$.
The threshold wavelength $\lambda_0$ is calculated as:
$\lambda_0 = \frac{1240 \ eV \cdot nm}{\phi_0} = \frac{1240}{2.4} \approx 516.7 \ nm$.
Photoemission will not take place if the incident wavelength $\lambda > \lambda_0$.
Comparing the given options:
$A) 200 \ nm < 516.7 \ nm$ (Emission occurs)
$B) 300 \ nm < 516.7 \ nm$ (Emission occurs)
$C) 700 \ nm > 516.7 \ nm$ (No emission)
$D) 400 \ nm < 516.7 \ nm$ (Emission occurs)
Therefore,for $700 \ nm$,photoemission does not take place.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by:
$K_{max} = h\nu - \phi_0$
Since $K_{max} = eV_0$,where $V_0$ is the stopping potential,we have:
$eV_0 = h\nu - \phi_0$
Dividing by $e$,we get:
$V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$
Comparing this equation with the straight-line equation $y = mx + c$,where $y = V_0$ and $x = \nu$:
Slope $(m)$ = $\frac{h}{e}$
$y$-intercept $(c)$ = $-\frac{\phi_0}{e}$
Since the work function $\phi_0 = h\nu_0$,where $\nu_0$ is the threshold frequency,the $y$-intercept becomes:
$c = -\frac{h\nu_0}{e}$
Thus,the slope is $\frac{h}{e}$ and the $y$-intercept is $-\frac{h\nu_0}{e}$.

(A) Given,the change in kinetic energy of photoelectrons is $\Delta K = 0.52 eV$. The initial wavelength is $\lambda_1 = 500 \,nm$.
Using Einstein's photoelectric equation,$K = \frac{hc}{\lambda} - \phi$.
For two different wavelengths $\lambda_1$ and $\lambda_2$,the change in kinetic energy is given by:
$\Delta K = K_2 - K_1 = hc \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right)$.
Using $hc \approx 1242 \,eV \cdot nm$,we substitute the values:
$0.52 = 1242 \left( \frac{1}{\lambda_2} - \frac{1}{500} \right)$.
$\frac{0.52}{1242} = \frac{1}{\lambda_2} - 0.002$.
$0.000418 = \frac{1}{\lambda_2} - 0.002$.
$\frac{1}{\lambda_2} = 0.002418$.
$\lambda_2 = \frac{1}{0.002418} \approx 413.5 \,nm$.
Rounding to the nearest option,the new wavelength is approximately $400 \,nm$.

(D) Given: Work-function, $\phi_{0} = 1 eV = 1.6 \times 10^{-19} J$.
Wavelength, $\lambda = 3000 \text{Å} = 3000 \times 10^{-10} m$.
The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3000 \times 10^{-10}} = 6.63 \times 10^{-19} J$.
According to Einstein's photoelectric equation, $E = \phi_{0} + KE_{max}$.
$KE_{max} = E - \phi_{0} = 6.63 \times 10^{-19} - 1.6 \times 10^{-19} = 5.03 \times 10^{-19} J$.
Since $KE_{max} = \frac{1}{2}mv^{2}$, we have $v = \sqrt{\frac{2 \times KE_{max}}{m}}$.
$v = \sqrt{\frac{2 \times 5.03 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{1.105 \times 10^{12}} \approx 1.05 \times 10^{6} m/s$.
Thus, the velocity is approximately $1 \times 10^{6} m/s$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of an ejected photoelectron 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.
Since $\phi_0 = h\nu_0$,where $\nu_0$ is the threshold frequency,we can write:
$K_{max} = h\nu - h\nu_0 = h(\nu - \nu_0)$
This equation represents a straight line of the form $y = mx + c$,where the slope is $h$ and the intercept on the frequency axis is $\nu_0$.
For frequencies $\nu < \nu_0$,the kinetic energy is zero because no photoemission occurs. For $\nu \geq \nu_0$,the kinetic energy increases linearly with frequency. This behavior is correctly represented by graph $D$.

(B) In the photoelectric effect,the maximum kinetic energy of emitted photoelectrons depends on the frequency of the incident radiation.
According to Einstein's photoelectric equation:
$KE_{\max} = h\nu - \phi$
Where:
$KE_{\max}$ is the maximum kinetic energy of the emitted photoelectrons.
$h$ is Planck's constant.
$\nu$ is the frequency of the incident radiation.
$\phi$ is the work function of the metal.
Since the work function $\phi$ is a constant for a given metal,the maximum kinetic energy $(KE_{\max})$ is directly dependent on the frequency $\nu$ of the incident radiation.

(C) According to Einstein's photoelectric equation,the stopping potential $V_{0}$ is related to the frequency of incident radiation $v$ by the relation: $eV_{0} = h\nu - \phi$,where $h$ is Planck's constant and $\phi$ is the work function of the metal.
From the graph,the stopping potentials for the three frequencies are $V_{01}, V_{02},$ and $V_{03}$ respectively.
It is observed that $|V_{03}| > |V_{02}| > |V_{01}|$.
Since the stopping potential is directly proportional to the frequency of the incident radiation,a higher magnitude of stopping potential corresponds to a higher frequency.
Therefore,the relationship between the frequencies is $v_{3} > v_{2} > v_{1}$,which is equivalent to $v_{1} < v_{2} < v_{3}$.

(D) The energy of a photon emitted during a transition from $n_i$ to $n_f$ is given by $E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
For the transition $n=3$ to $n=2$, the energy is $E_{3-2} = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{5}{36} \right) \approx 1.89 \text{ eV}$.
Since photoelectrons are just emitted, the work function of the metal is $\Phi = 1.89 \text{ eV}$.
For the photoelectric effect to occur, the energy of the incident photon must be greater than or equal to the work function $(E \ge \Phi)$.
Let's calculate the energy for the given options:
$A$: $n=2$ to $n=1$: $E = 13.6 (1 - 1/4) = 10.2 \text{ eV} > 1.89 \text{ eV}$ (Possible).
$B$: $n=3$ to $n=1$: $E = 13.6 (1 - 1/9) = 12.09 \text{ eV} > 1.89 \text{ eV}$ (Possible).
$C$: $n=5$ to $n=2$: $E = 13.6 (1/4 - 1/25) = 13.6 (21/100) = 2.856 \text{ eV} > 1.89 \text{ eV}$ (Possible).
$D$: $n=4$ to $n=3$: $E = 13.6 (1/9 - 1/16) = 13.6 (7/144) \approx 0.66 \text{ eV} < 1.89 \text{ eV}$ (Not possible).
Therefore, the photoelectric effect is not possible for the transition from $n=4$ to $n=3$.

(C) Given: Maximum velocity $v = 1.8 \times 10^{6} \ m/s$ and specific charge $\frac{e}{m} = 1.8 \times 10^{11} \ C/kg$.
The kinetic energy of the fastest photoelectron is equal to the work done by the stopping potential $V_{0}$,given by the equation: $e V_{0} = \frac{1}{2} m v^{2}$.
Dividing both sides by $m$,we get: $V_{0} \left(\frac{e}{m}\right) = \frac{v^{2}}{2}$.
Substituting the given values:
$V_{0} \times (1.8 \times 10^{11}) = \frac{(1.8 \times 10^{6})^{2}}{2}$.
$V_{0} \times 1.8 \times 10^{11} = \frac{3.24 \times 10^{12}}{2}$.
$V_{0} \times 1.8 \times 10^{11} = 1.62 \times 10^{12}$.
$V_{0} = \frac{1.62 \times 10^{12}}{1.8 \times 10^{11}} = 0.9 \times 10 = 9 \ V$.

(D) According to Einstein's photoelectric equation,the energy of the incident photon is equal to the sum of the work function $(W)$ and the maximum kinetic energy $(KE)$ of the ejected electron.
$E = W + KE$
$KE = E - W$
Since $E = \frac{hc}{\lambda}$ and $W = \frac{hc}{\lambda_{0}}$,we have:
$KE = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}}$
$KE = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$
$KE = hc \left( \frac{\lambda_{0} - \lambda}{\lambda \lambda_{0}} \right)$

(C) According to Einstein's photoelectric equation:
$KE_{max} = h\nu - \phi_{0}$
Comparing this with the equation of a straight line $y = mx + c$,where $y = KE_{max}$,$x = \nu$,$m$ is the slope,and $c$ is the y-intercept:
$KE_{max} = h\nu + (-\phi_{0})$
Here,the slope $m = h$ (Planck's constant).
Since $h$ is a universal constant,the slope is the same for all metals and is independent of the intensity of the incident radiation.

(A) According to Einstein's photoelectric equation: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$.
For the first case: $3 eV_{s} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$ --- $(i)$
For the second case: $eV_{s} = hc \left( \frac{1}{2 \lambda} - \frac{1}{\lambda_{0}} \right)$ --- (ii)
Dividing equation $(i)$ by equation (ii):
$\frac{3 eV_{s}}{eV_{s}} = \frac{hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)}{hc \left( \frac{1}{2 \lambda} - \frac{1}{\lambda_{0}} \right)}$
$3 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_{0}}}{\frac{1}{2 \lambda} - \frac{1}{\lambda_{0}}}$
$3 \left( \frac{1}{2 \lambda} - \frac{1}{\lambda_{0}} \right) = \frac{1}{\lambda} - \frac{1}{\lambda_{0}}$
$\frac{3}{2 \lambda} - \frac{3}{\lambda_{0}} = \frac{1}{\lambda} - \frac{1}{\lambda_{0}}$
$\frac{3}{2 \lambda} - \frac{1}{\lambda} = \frac{3}{\lambda_{0}} - \frac{1}{\lambda_{0}}$
$\frac{1}{2 \lambda} = \frac{2}{\lambda_{0}}$
$\lambda_{0} = 4 \lambda$.

(C) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ of emitted electrons 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 of the metal.
Since $\nu = c / \lambda$, the kinetic energy depends on the frequency $\nu$ and the wavelength $\lambda$ of the incident light, as well as the work function $\Phi_0$ of the material.
The intensity of incident light determines the number of photons striking the surface per unit time, which affects the number of emitted electrons (photoelectric current), but it does not affect the maximum kinetic energy of individual emitted electrons.

(D) According to Einstein's photoelectric equation:
$E_{k} = h\nu - \Phi$
where $E_{k}$ is the maximum kinetic energy,$h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\Phi = h\nu_{0}$ is the work function of the metal surface ($\nu_{0}$ is the threshold frequency).
This equation is of the form $y = mx + c$,where:
$y = E_{k}$
$x = \nu$
$m = h$ (slope)
$c = -\Phi$ (y-intercept)
$1$. For $\nu < \nu_{0}$,the energy of the incident photon is less than the work function,so no photoelectric emission occurs,and $E_{k} = 0$.
$2$. For $\nu = \nu_{0}$,$E_{k} = 0$.
$3$. For $\nu > \nu_{0}$,$E_{k}$ increases linearly with frequency $\nu$.
Graph $D$ correctly shows that $E_{k}$ is zero until the threshold frequency $\nu_{0}$ is reached,after which it increases linearly.

(B) The threshold energy (work function) is given by $\Phi = h \nu_0$. Given $h = 6.6 \times 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 \times 10^{-19} \ J$.
For metal $A$:
$\Phi_A = h \nu_A = (6.6 \times 10^{-34}) \times (1.8 \times 10^{14}) \ J = 11.88 \times 10^{-20} \ J$.
Converting to $eV$: $\Phi_A = \frac{11.88 \times 10^{-20}}{1.6 \times 10^{-19}} \ eV = 0.7425 \ eV$.
For metal $B$:
$\Phi_B = h \nu_B = (6.6 \times 10^{-34}) \times (2.2 \times 10^{14}) \ J = 14.52 \times 10^{-20} \ J$.
Converting to $eV$: $\Phi_B = \frac{14.52 \times 10^{-20}}{1.6 \times 10^{-19}} \ eV = 0.9075 \ eV$.
The energy of incident photons is $E = 0.825 \ eV$.
Since $E > \Phi_A$ $(0.825 \ eV > 0.7425 \ eV)$ and $E < \Phi_B$ $(0.825 \ eV < 0.9075 \ eV)$,photoelectrons are emitted only from metal $A$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted electrons 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.
For the first photon with energy $E_1 = 1 \text{ eV}$:
$K_1 = 1 \text{ eV} - 0.5 \text{ eV} = 0.5 \text{ eV}$.
For the second photon with energy $E_2 = 2.5 \text{ eV}$:
$K_2 = 2.5 \text{ eV} - 0.5 \text{ eV} = 2.0 \text{ eV}$.
Since $K_{max} = \frac{1}{2}mv^2$,the ratio of the kinetic energies is:
$\frac{K_1}{K_2} = \frac{0.5}{2.0} = \frac{1}{4}$.
Substituting the expression for kinetic energy:
$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{1}{4} \Rightarrow \frac{v_1^2}{v_2^2} = \frac{1}{4}$.
Taking the square root on both sides:
$\frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio of the maximum speeds is $1:2$.

(B) When light of suitable frequency falls on a metal plate,electrons are emitted. This phenomenon is known as the photoelectric effect.
According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{\max})$ of the emitted photo-electrons is given by:
$K_{\max} = hf - \phi_0$
where $h$ is Planck's constant,$f$ is the frequency of the incident light,and $\phi_0$ is the work function of the metal.
Since $h$ and $\phi_0$ are constants for a given metal,the maximum kinetic energy depends only on the frequency $(f)$ of the incident light.

(D) In the Paschen series,an electron transitions from a higher energy state to the $n=3$ state. The energy of a photon in the Paschen series is given by $E = 13.6 \left( \frac{1}{3^2} - \frac{1}{n^2} \right) \text{ eV}$,where $n = 4, 5, 6, \dots$
The minimum energy corresponds to the transition from $n=4$ to $n=3$:
$E_{\min} = 13.6 \left( \frac{1}{9} - \frac{1}{16} \right) = 13.6 \left( \frac{16-9}{144} \right) = 13.6 \left( \frac{7}{144} \right) \approx 0.66 \text{ eV}$.
The maximum energy corresponds to the transition from $n=\infty$ to $n=3$:
$E_{\max} = 13.6 \left( \frac{1}{9} - 0 \right) = \frac{13.6}{9} \approx 1.51 \text{ eV}$.
For a photon to eject a photoelectron,its energy must be greater than or equal to the work function $\phi_0$ of the metal. Thus,$\phi_0 \leq E_{\text{photon}}$. Since the maximum energy of a Paschen photon is $1.51 \text{ eV}$,the work function must satisfy $\phi_0 \leq 1.51 \text{ eV}$.
Among the given options,$1.1 \text{ eV}$ is the only value that satisfies this condition.

(B) The work function $\Phi$ is given by the formula $\Phi = \frac{hc}{\lambda_0}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda_0$ is the threshold wavelength.
Given: $h = 6.6 \times 10^{-34} \ \text{Js}$, $c = 3 \times 10^8 \ \text{m/s}$, and $\lambda_0 = 6250 \ \text{Å} = 6250 \times 10^{-10} \ \text{m}$.
Substituting the values:
$\Phi = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6250 \times 10^{-10}} \ \text{J}$.
$\Phi = \frac{19.8 \times 10^{-26}}{6250 \times 10^{-10}} = \frac{19.8}{6250} \times 10^{-16} \ \text{J} = 3.168 \times 10^{-20} \ \text{J}$.
To convert this into electron-volts (eV), divide by the charge of an electron $e = 1.6 \times 10^{-19} \ \text{C}$:
$\Phi = \frac{3.168 \times 10^{-20}}{1.6 \times 10^{-19}} \ \text{eV} = 1.98 \ \text{eV}$.
Thus, the correct option is $B$.

(A) The energy of the incident photon is $E = 8 \times 10^{-19} \ J$.
Converting this to electron-volts,$E = \frac{8 \times 10^{-19}}{1.6 \times 10^{-19}} \ eV = 5 \ eV$.
The de Broglie wavelength $\lambda$ of the emitted photoelectrons is $10 \ Å = 10^{-9} \ m$.
The kinetic energy $K_{max}$ of the photoelectrons is given by $K_{max} = \frac{h^2}{2m\lambda^2}$.
Substituting the values: $K_{max} = \frac{(6.63 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (10^{-9})^2} \approx 2.41 \times 10^{-19} \ J$.
Converting to $eV$: $K_{max} = \frac{2.41 \times 10^{-19}}{1.6 \times 10^{-19}} \ eV \approx 1.5 \ eV$.
Using Einstein's photoelectric equation: $E = \phi + K_{max}$,where $\phi$ is the work function.
$\phi = E - K_{max} = 5 \ eV - 1.5 \ eV = 3.5 \ eV$.

(A) Given: Work function $\phi_0 = 4.2 \text{ eV}$,Wavelength $\lambda = 2000 \text{ Å}$.
Energy of incident photon $E = \frac{12400}{\lambda(\text{in Å})} \text{ eV} = \frac{12400}{2000} = 6.2 \text{ eV}$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max} = E - \phi_0$.
$K_{\max} = 6.2 \text{ eV} - 4.2 \text{ eV} = 2 \text{ eV}$.
Converting kinetic energy to Joules: $K_{\max} = 2 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-19} \text{ J}$.
Using $K_{\max} = \frac{1}{2} m v_{\max}^2$,where mass of electron $m = 9.1 \times 10^{-31} \text{ kg}$.
$v_{\max}^2 = \frac{2 \times K_{\max}}{m} = \frac{2 \times 3.2 \times 10^{-19}}{9.1 \times 10^{-31}} \approx 0.703 \times 10^{12} \text{ m}^2/\text{s}^2$.
$v_{\max} = \sqrt{0.703 \times 10^{12}} \approx 8.38 \times 10^5 \text{ m/s} \approx 8.4 \times 10^5 \text{ m/s}$.

(A) The work function $\phi_0$ is given as $9 \ eV$.
The threshold wavelength $\lambda_0$ is the longest wavelength capable of initiating the photoelectric effect.
The relationship between work function and threshold wavelength is given by $\phi_0 = \frac{hc}{\lambda_0}$.
Using the constant $hc \approx 1240 \ eV \cdot nm$,we have:
$\lambda_0 = \frac{1240 \ eV \cdot nm}{9 \ eV} \approx 137.77 \ nm$.
Converting nanometers to meters:
$\lambda_0 \approx 137.77 \times 10^{-9} \ m = 1.3777 \times 10^{-7} \ m$.
Rounding to the nearest option,we get $1.37 \times 10^{-7} \ m$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:
$K_{\max} = \frac{hc}{\lambda} - \phi_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$
Since $K_{\max} = \frac{1}{2}mv_{\max}^2$,we have:
$\frac{1}{2}mv_{\max}^2 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$
$v_{\max} = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)}$
For the first beam with $\lambda_1 = \frac{\lambda_0}{3}$:
$v_1 = \sqrt{\frac{2hc}{m} \left( \frac{3}{\lambda_0} - \frac{1}{\lambda_0} \right)} = \sqrt{\frac{2hc}{m} \cdot \frac{2}{\lambda_0}}$
For the second beam with $\lambda_2 = \frac{\lambda_0}{9}$:
$v_2 = \sqrt{\frac{2hc}{m} \left( \frac{9}{\lambda_0} - \frac{1}{\lambda_0} \right)} = \sqrt{\frac{2hc}{m} \cdot \frac{8}{\lambda_0}}$
The ratio of the maximum velocities is:
$\frac{v_1}{v_2} = \sqrt{\frac{2/\lambda_0}{8/\lambda_0}} = \sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

(D) Given: $\lambda_1 = 300 \ nm$,$\lambda_2 = 500 \ nm$,and the ratio of maximum velocities $\frac{v_1}{v_2} = 3$.
Using Einstein's photoelectric equation: $E = \phi_0 + K_{max}$,where $K_{max} = \frac{1}{2}mv^2$.
Thus,$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi_0$.
Taking the ratio for the two cases:
$\left(\frac{v_1}{v_2}\right)^2 = \frac{E_1 - \phi_0}{E_2 - \phi_0} = \frac{\frac{1240}{\lambda_1} - \phi_0}{\frac{1240}{\lambda_2} - \phi_0}$.
Substituting the values: $3^2 = \frac{\frac{1240}{300} - \phi_0}{\frac{1240}{500} - \phi_0}$.
$9 = \frac{4.133 - \phi_0}{2.48 - \phi_0}$.
$9(2.48 - \phi_0) = 4.133 - \phi_0$.
$22.32 - 9\phi_0 = 4.133 - \phi_0$.
$8\phi_0 = 18.187$.
$\phi_0 \approx 2.273 \ eV \approx 2.28 \ eV$.

(B) Given:
Work function,$\phi_0 = 2 \text{ eV}$
Maximum kinetic energy,$K_{\max} = 2 \text{ eV}$
According to Einstein's photoelectric equation:
$E = \phi_0 + K_{\max}$
$E = 2 \text{ eV} + 2 \text{ eV} = 4 \text{ eV}$
The energy of a photon is given by $E = \frac{hc}{\lambda}$.
Using the relation $\lambda = \frac{12400 \text{ eV Å}}{E \text{ (in eV)}}$:
$\lambda = \frac{12400}{4} \text{ Å} = 3100 \text{ Å}$
Thus,the wavelength of the incident photon is $3100 \text{ Å}$.

(C) According to Einstein's photoelectric equation:
$h \nu = \phi + K.E._{max}$
$h \nu = \phi + \frac{1}{2} m v^2$
We know that the de Broglie wavelength is given by $\lambda = \frac{h}{mv}$,which implies $v = \frac{h}{m \lambda}$.
Substituting the value of $v$ into the kinetic energy equation:
$h \nu = \phi + \frac{1}{2} m \left( \frac{h}{m \lambda} \right)^2$
$h \nu = \phi + \frac{1}{2} m \left( \frac{h^2}{m^2 \lambda^2} \right)$
$h \nu = \phi + \frac{h^2}{2 m \lambda^2}$
Dividing both sides by $h$:
$\nu = \frac{\phi}{h} + \frac{h}{2 m \lambda^2}$

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = h\nu - \phi$.
Given that for frequency $\nu$,the maximum kinetic energy is $E$,we have:
$E = h\nu - \phi$ --- $(1)$
When the frequency is increased to $3\nu$,the new maximum kinetic energy $K'_{max}$ is:
$K'_{max} = h(3\nu) - \phi = 3h\nu - \phi$ --- $(2)$
From equation $(1)$,we can express $h\nu$ as $h\nu = E + \phi$.
Substituting this into equation $(2)$:
$K'_{max} = 3(E + \phi) - \phi$
$K'_{max} = 3E + 3\phi - \phi$
$K'_{max} = 3E + 2\phi$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = h\nu - h\nu_0$,where $\nu_0$ is the threshold frequency.
Since $K_{max} = \frac{P^2}{2m}$,we have $\frac{P^2}{2m} = h\nu - h\nu_0$.
For the first case,$\nu = 2v$ and $\nu_0 = v$:
$\frac{P^2}{2m} = h(2v) - hv = hv$ ... $(1)$
For the second case,let the new frequency be $\nu'$ and the new momentum be $2P$:
$\frac{(2P)^2}{2m} = h\nu' - hv$
$\frac{4P^2}{2m} = h\nu' - hv$ ... $(2)$
Substituting the value of $\frac{P^2}{2m}$ from equation $(1)$ into equation $(2)$:
$4(hv) = h\nu' - hv$
$4hv + hv = h\nu'$
$h\nu' = 5hv$
$\nu' = 5v$

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $E$ is given by: $E = \frac{hc}{\lambda} - \phi$ ... $(i)$,
where $\phi$ is the work function.
When the wavelength is reduced by $50 \%$,the new wavelength $\lambda' = \lambda - 0.5\lambda = 0.5\lambda = \frac{\lambda}{2}$.
The new maximum kinetic energy $E' = 3E$.
Substituting these into the photoelectric equation: $3E = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$ ... $(ii)$.
From equation $(i)$,we have $E = \frac{hc}{\lambda} - \phi$. Substituting this into equation $(ii)$:
$3(\frac{hc}{\lambda} - \phi) = \frac{2hc}{\lambda} - \phi$
$\frac{3hc}{\lambda} - 3\phi = \frac{2hc}{\lambda} - \phi$
$\frac{3hc}{\lambda} - \frac{2hc}{\lambda} = 3\phi - \phi$
$\frac{hc}{\lambda} = 2\phi$
$\phi = \frac{hc}{2\lambda}$.

(B) The photoelectric equation is given by $eV_s = h\nu - \phi$,where $V_s$ is the stopping potential,$h$ is Planck's constant,$\nu$ is the frequency,and $\phi$ is the work function.
Rearranging the equation for $V_s$ as a function of $\nu$: $V_s = (\frac{h}{e})\nu - \frac{\phi}{e}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = V_s$ and $x = \nu$,the slope of the graph with respect to the $X$-axis is $m = \frac{h}{e}$.
However,the question specifies that the angle $\theta$ is made with the $Y$-axis. For a line $y = mx + c$,the slope $m = \tan(\alpha)$ where $\alpha$ is the angle with the $X$-axis. The angle $\theta$ with the $Y$-axis is related to the angle $\alpha$ with the $X$-axis by $\theta = 90^\circ - \alpha$.
Therefore,$\tan \theta = \tan(90^\circ - \alpha) = \cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{h/e} = \frac{e}{h}$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{max} = hf - \phi$.
For the first case: $f_1 = 3f$,$\phi = 2hf$,and $v_1 = v$.
$\frac{1}{2}mv^2 = 3hf - 2hf = hf$ --- $(1)$
For the second case: $f_2 = 4.25f$,$\phi = 2hf$,and $v_2 = ?$.
$\frac{1}{2}mv_2^2 = 4.25hf - 2hf = 2.25hf$ --- $(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{\frac{1}{2}mv_2^2}{\frac{1}{2}mv^2} = \frac{2.25hf}{hf}$
$\frac{v_2^2}{v^2} = 2.25$
$v_2^2 = 2.25v^2$
$v_2 = \sqrt{2.25}v = 1.5v$.

(C) The frequency of the $H_{\alpha}$ line of the Balmer series is given by $\nu_0 = Rc(\frac{1}{2^2} - \frac{1}{3^2}) = Rc(\frac{1}{4} - \frac{1}{9}) = \frac{5Rc}{36}$. This is the threshold frequency $\nu_0$ of the material.
The frequency of the $H_{\beta}$ line of the Balmer series is given by $\nu = Rc(\frac{1}{2^2} - \frac{1}{4^2}) = Rc(\frac{1}{4} - \frac{1}{16}) = \frac{3Rc}{16}$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = h\nu - h\nu_0$.
Substituting the values:
$K_{max} = h(\frac{3Rc}{16}) - h(\frac{5Rc}{36})$
$K_{max} = Rhc(\frac{3}{16} - \frac{5}{36})$
$K_{max} = Rhc(\frac{27 - 20}{144}) = \frac{7 Rhc}{144}$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy of the 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 of the metal.
This equation shows that $K_{max}$ depends only on the frequency of the incident radiation and is independent of the intensity of the incident radiation.
Therefore,option $(C)$ is correct.
For a given frequency,the photocurrent is directly proportional to the intensity of incident radiation.
The stopping potential depends only on the frequency of incident radiation and is independent of the intensity.
If the frequency of incident radiation is less than the threshold (cutoff) frequency,no photoelectric emission occurs,regardless of how much the intensity is increased.

(C) Given the wave equation: $E = E_0 \sin [k(x - ct)]$,where $k = \frac{2\pi}{\lambda} = 1.57 \times 10^7 \text{ m}^{-1}$.
Calculating wavelength $\lambda$: $\lambda = \frac{2 \times 3.14}{1.57 \times 10^7} = 4 \times 10^{-7} \text{ m}$.
Energy of incident photon $E_p = \frac{hc}{\lambda} = \frac{6.64 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} = 4.98 \times 10^{-19} \text{ J}$.
Converting work function $\phi_0$ to Joules: $\phi_0 = 1.9 \text{ eV} = 1.9 \times 1.6 \times 10^{-19} \text{ J} = 3.04 \times 10^{-19} \text{ J}$.
Using Einstein's photoelectric equation: $E_p = \phi_0 + eV_0$.
$eV_0 = 4.98 \times 10^{-19} - 3.04 \times 10^{-19} = 1.94 \times 10^{-19} \text{ J}$.
$V_0 = \frac{1.94 \times 10^{-19}}{1.6 \times 10^{-19}} \text{ V} = 1.2125 \text{ V}$.
Rounding to the nearest option,the stopping potential is $1.1 \text{ V}$.

(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi = \frac{1}{2}mv^2$.
For $\lambda_1 = 200 \,nm$,let the velocity be $v$. Then $\frac{hc}{200 \times 10^{-9}} - \phi = \frac{1}{2}mv^2$ $(i)$.
For $\lambda_2 = 600 \,nm$,the velocity is $\frac{v}{3}$. Then $\frac{hc}{600 \times 10^{-9}} - \phi = \frac{1}{2}m(\frac{v}{3})^2 = \frac{1}{9}(\frac{1}{2}mv^2)$ (ii).
From $(i)$,$\frac{1}{2}mv^2 = \frac{hc}{200 \times 10^{-9}} - \phi$.
Substitute this into (ii): $\frac{hc}{600 \times 10^{-9}} - \phi = \frac{1}{9}(\frac{hc}{200 \times 10^{-9}} - \phi)$.
Multiply by $9$: $\frac{9hc}{600 \times 10^{-9}} - 9\phi = \frac{hc}{200 \times 10^{-9}} - \phi$.
$8\phi = hc(\frac{9}{600 \times 10^{-9}} - \frac{1}{200 \times 10^{-9}}) = hc(\frac{9-3}{600 \times 10^{-9}}) = hc(\frac{6}{600 \times 10^{-9}}) = hc(\frac{1}{100 \times 10^{-9}}) = hc \times 10^7$.
Therefore,$\phi = \frac{hc}{8} \times 10^7 \,J$.

(C) Given: Work function $\phi_0 = 2.5 eV$,Frequency $f = 3.2 \times 10^{15} Hz$,Planck's constant $h = 6.6 \times 10^{-34} J-s$.
Energy of incident photon $E = hf$.
$E = (6.6 \times 10^{-34} J-s) \times (3.2 \times 10^{15} Hz) = 21.12 \times 10^{-19} J$.
To convert energy into $eV$,divide by $1.6 \times 10^{-19} J/eV$:
$E = \frac{21.12 \times 10^{-19}}{1.6 \times 10^{-19}} eV = 13.2 eV$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max} = E - \phi_0$.
$K_{max} = 13.2 eV - 2.5 eV = 10.7 eV$.

(B) The intensity $I$ is given as $100 \ W \ m^{-2}$. The energy of a single photon is $E = \frac{hc}{\lambda}$.
Number of photons incident per unit area per unit time $(N)$ is given by $N = \frac{I}{E} = \frac{I \lambda}{hc}$.
Substituting the values: $N = \frac{100 \times 300 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8} \approx 1.51 \times 10^{20} \ m^{-2} \ s^{-1}$.
Given that $2 \%$ of incident photons produce photoelectrons,the number of photoelectrons emitted per unit area per unit time is $n' = 0.02 \times N = 0.02 \times 1.51 \times 10^{20} = 3.02 \times 10^{18} \ m^{-2} \ s^{-1}$.
For an area $A = 2 \ cm^2 = 2 \times 10^{-4} \ m^2$,the number of photoelectrons emitted per second is $n = n' \times A = 3.02 \times 10^{18} \times 2 \times 10^{-4} = 6.04 \times 10^{14}$.

(C) According to Einstein's photoelectric equation:
$h\nu = h\nu_0 + \frac{1}{2}mv_{\text{max}}^2$
Where $h$ is Planck's constant,$\nu$ is the frequency of incident radiation,$\nu_0$ is the threshold frequency,$m$ is the mass of the electron,and $v_{\text{max}}$ is the maximum speed of the photoelectron.
Rearranging the equation for $v_{\text{max}}$:
$\frac{1}{2}mv_{\text{max}}^2 = h(\nu - \nu_0)$
$v_{\text{max}}^2 = \frac{2h}{m}(\nu - \nu_0)$
$v_{\text{max}} = \sqrt{\frac{2h}{m}} \sqrt{\nu - \nu_0}$
For $\nu < \nu_0$,$v_{\text{max}} = 0$. For $\nu \ge \nu_0$,$v_{\text{max}}$ increases with $\nu$. The relationship is $v_{\text{max}} \propto \sqrt{\nu - \nu_0}$. This represents a parabolic curve starting from the threshold frequency $\nu_0$ on the frequency axis,where the slope decreases as $\nu$ increases. Thus,option $C$ is the correct representation.

(B) The Einstein's photoelectric equation is given by: $K E_{\max} = \frac{hc}{\lambda} - W$,where $W$ is the work function and $\lambda$ is the wavelength of incident light.
Given $\lambda = 1240 \mathring{A} = 1240 \times 10^{-10} \text{ m}$ and stopping potential $V_0 = 8 \text{ V}$.
The energy of incident photon is $E = \frac{hc}{\lambda} = \frac{12400 \text{ eV} \cdot \mathring{A}}{1240 \mathring{A}} = 10 \text{ eV}$.
Since $K E_{\max} = eV_0 = 8 \text{ eV}$,we have $8 \text{ eV} = 10 \text{ eV} - W$.
Therefore,the work function $W = 10 \text{ eV} - 8 \text{ eV} = 2 \text{ eV}$.
The threshold wavelength $\lambda_0$ is given by $W = \frac{hc}{\lambda_0}$.
$\lambda_0 = \frac{12400 \text{ eV} \cdot \mathring{A}}{W} = \frac{12400 \text{ eV} \cdot \mathring{A}}{2 \text{ eV}} = 6200 \mathring{A}$.

(B) In the study of the photoelectric effect,light energy is considered to exist in the form of discrete packets. Each packet of this energy is known as a photon.
The energy of one photon is given by $E = h \nu$,where $h$ is Planck's constant and $\nu$ is the frequency of the photon.
Since the energy is restricted to discrete values (multiples of $h\nu$),the study of the photoelectric effect is fundamental in understanding the quantisation of energy.

(C) Given, wavelength of light, $\lambda = 4900 Å$.
Stopping potential, $V_s = 2 \,V$.
The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using the approximation $E \approx \frac{12400}{\lambda (Å)} eV$:
$E = \frac{12400}{4900} eV \approx 2.53 eV$.
The maximum kinetic energy of the emitted electrons is $K_{max} = e V_s = 2 eV$.
According to Einstein's photoelectric equation, $K_{max} = E - \phi$, where $\phi$ is the work function.
Substituting the values: $2 eV = 2.53 eV - \phi$.
Therefore, $\phi = 2.53 eV - 2 eV = 0.53 eV$.

(A) Energy of the incident ultraviolet radiation,$E = 6.2 eV$.
Work-function of the aluminium surface,$\phi = 4.2 eV$.
According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{\max})$ of the emitted electron is given by:
$K_{\max} = E - \phi$
$K_{\max} = 6.2 eV - 4.2 eV = 2.0 eV$.
To convert this energy into Joules,we use the conversion factor $1 eV = 1.6 \times 10^{-19} J$:
$K_{\max} = 2.0 \times 1.6 \times 10^{-19} J = 3.2 \times 10^{-19} J$.

(C) Given: Work function $\phi_0 = 1.24 eV$.
Wavelength $\lambda = 4.36 \times 10^{-7} m$.
Using the relation $E = \frac{hc}{\lambda}$,where $hc \approx 1240 eV \cdot nm = 1240 \times 10^{-9} eV \cdot m$.
$E = \frac{1240 \times 10^{-9} eV \cdot m}{4.36 \times 10^{-7} m} \approx 2.844 eV$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max} = E - \phi_0$.
$K_{max} = 2.844 eV - 1.24 eV = 1.604 eV$.
Since $K_{max} = eV_0$,the retarding potential $V_0$ is $1.604 V \approx 1.60 V$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy of an emitted electron is given by:
$K_{\max} = \frac{hc}{\lambda} - \phi$
Since $K_{\max} = \frac{1}{2}mv_{\max}^2$,we have:
$\frac{1}{2}mv_{\max}^2 = \frac{hc}{\lambda} - \phi$
$\frac{1}{2}mv_{\max}^2 = \frac{hc - \lambda\phi}{\lambda}$
$v_{\max}^2 = \frac{2(hc - \lambda\phi)}{m\lambda}$
$v_{\max} = \sqrt{\frac{2(hc - \lambda\phi)}{m\lambda}}$

(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.
This equation shows that $K_{max}$ depends only on the frequency of the incident light and the nature of the metal surface.
Intensity of light affects the number of photons incident per unit area per unit time, which in turn affects the number of photoelectrons emitted (photoelectric current), but it does not affect the energy of individual photoelectrons.
Therefore, when the intensity is reduced to one-fourth, the maximum kinetic energy of the emitted photoelectrons remains unchanged.