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

(C) The maximum kinetic energy of photoelectrons is given by Einstein's photoelectric equation: $K_{\max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
Since $K_{\max} = eV_s$,we have $V_s = \frac{hc}{e\lambda} - \frac{\phi}{e}$.
For $\lambda_1 = 4000 \, \mathring{A}$,$V_{s1} = \frac{12400}{4000} - \frac{\phi}{e} = 3.1 - \frac{\phi}{e}$ (using $hc \approx 12400 \, \text{eV} \mathring{A}$).
For $\lambda_2 = 3600 \, \mathring{A}$,$V_{s2} = \frac{12400}{3600} - \frac{\phi}{e} \approx 3.444 - \frac{\phi}{e}$.
The change in stopping potential is $\Delta V_s = V_{s2} - V_{s1} = (3.444 - \frac{\phi}{e}) - (3.1 - \frac{\phi}{e}) = 3.444 - 3.1 = 0.344 \, V \approx 0.34 \, V$.

(D) $1$. Saturation current $(i_s)$ is directly proportional to the intensity of incident light. Since the intensity is doubled, the saturation current becomes $2 \times 5\,mA = 10\,mA$.
$2$. Stopping potential $(V_s)$ is given by Einstein's photoelectric equation: $eV_s = h\nu - \Phi$, where $\Phi$ is the work function.
$3$. Initially, $eV_{s1} = h\nu - \Phi = 10\,eV$.
$4$. When frequency is doubled $(2\nu)$, the new stopping potential $V_{s2}$ satisfies $eV_{s2} = h(2\nu) - \Phi = 2h\nu - \Phi$.
$5$. Since $h\nu = 10 + \Phi$, we have $eV_{s2} = 2(10 + \Phi) - \Phi = 20 + 2\Phi - \Phi = 20 + \Phi$.
$6$. Since $\Phi > 0$, it follows that $V_{s2} > 20\,V$.
$7$. Therefore, $i_s = 10\,mA$ and $V_s > 20\,V$.

(B) The energy of the incident photon is $E = 4\,eV$.
The work function of the metal is $\phi = 2\,eV$.
For photoemission to occur,the energy of the photon must be greater than or equal to the work function plus the potential energy barrier created by the sphere's potential $V_0$.
For no photoelectrons to be ejected,the condition is $E < \phi + eV_0$.
Substituting the given values: $4\,eV < 2\,eV + eV_0$.
$2\,eV < eV_0$.
Therefore,$V_0 > 2\,V$.
The minimum potential required to stop the emission of photoelectrons is $2\,V$.

(C) 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 surface.
Photoelectrons are emitted from different depths within the metal surface.
Electrons emitted from the surface have the maximum kinetic energy $(K_{max})$.
Electrons emitted from deeper layers lose some energy due to collisions before escaping the surface.
Therefore, the kinetic energy of emitted photoelectrons ranges from $0$ to $K_{max}$.
Thus, the kinetic energy of any photoelectron is always less than or equal to the maximum kinetic energy $(K \le K_{max})$.

(C) According to Einstein's photoelectric equation,$KE_{\max} = hf - \Phi$,where $\Phi$ is the work function.
Statement $(A)$: Since $KE_{\max} = hf - \Phi$,if $f$ increases,$KE_{\max}$ increases. Thus,$(A)$ is correct.
Statement $(B)$: Stopping potential $V_s$ depends only on the frequency of incident light and the work function of the metal. It is independent of the distance between the cathode and anode. Thus,$(B)$ is incorrect.
Statement $(C)$: Stopping potential $V_s = \frac{KE_{\max}}{e} = \frac{hf - \Phi}{e}$. Since $V_s$ depends only on $f$ and $\Phi$,it remains constant if $I$ is increased. Saturation current is directly proportional to intensity $I$,so it increases. Thus,$(C)$ is correct.
Statement $(D)$: Since $V_s = \frac{hf - \Phi}{e}$,if $\Phi$ is decreased,$V_s$ increases. Thus,$(D)$ is correct.
Therefore,the correct statements are $(A), (C),$ and $(D)$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function of the metal.
For wavelength $\lambda_1$,$K_1 = \frac{hc}{\lambda_1} - \phi$.
For wavelength $\lambda_2$,$K_2 = \frac{hc}{\lambda_2} - \phi$.
Given $\lambda_1 = 2\lambda_2$,we substitute this into the expression for $K_1$:
$K_1 = \frac{hc}{2\lambda_2} - \phi$.
Since $\lambda_1 > \lambda_2$,the energy of the incident photon $\frac{hc}{\lambda_1}$ is less than $\frac{hc}{\lambda_2}$.
Therefore,$K_1 = \frac{hc}{2\lambda_2} - \phi < \frac{hc}{\lambda_2} - \phi = K_2$.
Thus,$K_1 < K_2$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy of an ejected electron is given by $K_{max} = \frac{1}{2} m_e v^2 = \frac{hc}{\lambda} - \Phi$,where $\Phi$ is the work function of the metal.
For wavelength ${\lambda _1}$:
$\frac{1}{2} m_e v_1^2 = \frac{hc}{\lambda_1} - \Phi$ --- $(1)$
For wavelength ${\lambda _2}$:
$\frac{1}{2} m_e v_2^2 = \frac{hc}{\lambda_2} - \Phi$ --- $(2)$
Subtracting equation $(1)$ from equation $(2)$:
$\frac{1}{2} m_e v_2^2 - \frac{1}{2} m_e v_1^2 = (\frac{hc}{\lambda_2} - \Phi) - (\frac{hc}{\lambda_1} - \Phi)$
$\frac{1}{2} m_e (v_2^2 - v_1^2) = hc (\frac{1}{\lambda_2} - \frac{1}{\lambda_1})$
Multiplying both sides by $\frac{2}{m_e}$:
$v_2^2 - v_1^2 = \frac{2hc}{m_e} [\frac{1}{\lambda_2} - \frac{1}{\lambda_1}]$

(D) According to Einstein's photoelectric equation:
$K_{\max} = h\nu - h\nu_0$
Since the maximum kinetic energy $K_{\max}$ must be positive for emission to occur,we have:
$h\nu - h\nu_0 > 0$
$h\nu > h\nu_0$
$\nu > \nu_0$
Therefore,photoelectric emission occurs only when the incident light has a frequency greater than a certain minimum frequency,known as the threshold frequency $(\nu_0)$.

(C) Einstein's photoelectric equation is given by $K_{max} = h\nu - \phi$, where $K_{max}$ is the maximum kinetic energy of the photoelectrons, $h$ 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 $y = K_{max}$, $x = \nu$, $m = h$ (slope), and $c = -\phi$ (y-intercept).
Since the slope $h$ is positive and the y-intercept $-\phi$ is negative, the graph is a straight line that starts from the x-axis at a threshold frequency $\nu_0 = \phi/h$ and has a positive slope.
Therefore, the graph starts from the x-axis at a positive frequency value.

(B) Let the work function of metal $A$ be $\phi_A = \phi_0$ and metal $B$ be $\phi_B = 2\phi_0$.
According to Einstein's photoelectric equation,the maximum kinetic energy $E_K$ is given by $E_K = h\nu - \phi$.
For metal $A$: $E_{K_A} = hf - \phi_0$.
For metal $B$: $E_{K_B} = h(2f) - 2\phi_0 = 2(hf - \phi_0)$.
Taking the ratio of the maximum kinetic energies:
$\frac{E_{K_A}}{E_{K_B}} = \frac{hf - \phi_0}{2(hf - \phi_0)} = \frac{1}{2}$.
Thus,the ratio is $1 : 2$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max}$ is given by:
$(KE)_{\max} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$
where $\lambda_0$ is the threshold wavelength.
For the initial case:
$(KE)_1 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$ .........$(i)$
When the wavelength is reduced to $\frac{\lambda}{3}$,the new kinetic energy is $n$ times the initial value:
$(KE)_2 = n(KE)_1 = hc \left( \frac{1}{\lambda/3} - \frac{1}{\lambda_0} \right) = hc \left( \frac{3}{\lambda} - \frac{1}{\lambda_0} \right)$ .........$(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$n = \frac{\frac{3}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{\lambda} - \frac{1}{\lambda_0}} = \frac{\frac{3\lambda_0 - \lambda}{\lambda \lambda_0}}{\frac{\lambda_0 - \lambda}{\lambda \lambda_0}} = \frac{3\lambda_0 - \lambda}{\lambda_0 - \lambda}$
$n(\lambda_0 - \lambda) = 3\lambda_0 - \lambda$
$n\lambda_0 - n\lambda = 3\lambda_0 - \lambda$
$n\lambda_0 - 3\lambda_0 = n\lambda - \lambda$
$\lambda_0(n - 3) = \lambda(n - 1)$
$\lambda_0 = \left( \frac{n - 1}{n - 3} \right) \lambda$

(C) The Einstein's photoelectric equation is given by $K_{max} = \frac{hc}{\lambda} - \phi$.
Initially,$5\,eV = \frac{hc}{\lambda} - \phi$ ..........$(1)$
When the wavelength is decreased by $60\%$,the new wavelength $\lambda' = \lambda - 0.6\lambda = 0.4\lambda = \frac{2}{5}\lambda$.
Substituting this into the equation for the new kinetic energy:
$17\,eV = \frac{hc}{0.4\lambda} - \phi = \frac{2.5hc}{\lambda} - \phi$ ..........$(2)$
From equation $(1)$,we have $\frac{hc}{\lambda} = 5\,eV + \phi$.
Substitute this into equation $(2)$:
$17\,eV = 2.5(5\,eV + \phi) - \phi$
$17\,eV = 12.5\,eV + 2.5\phi - \phi$
$17\,eV - 12.5\,eV = 1.5\phi$
$4.5\,eV = 1.5\phi$
$\phi = \frac{4.5}{1.5}\,eV = 3\,eV$.
Thus,the work function of the metal is $3\,eV$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
For the first case with wavelength $\lambda$: $K_1 = \frac{hc}{\lambda} - \phi$.
For the second case with wavelength $\lambda/2$: $K_2 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$.
Given that $K_2 = 3K_1$,we substitute the expressions:
$\frac{2hc}{\lambda} - \phi = 3 \left( \frac{hc}{\lambda} - \phi \right)$.
Expanding the equation: $\frac{2hc}{\lambda} - \phi = \frac{3hc}{\lambda} - 3\phi$.
Rearranging the terms to solve for $\phi$: $3\phi - \phi = \frac{3hc}{\lambda} - \frac{2hc}{\lambda}$.
$2\phi = \frac{hc}{\lambda}$.
Therefore,$\phi = \frac{hc}{2\lambda}$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $E_k$ of the emitted photoelectrons is given by:
$E_k = hv - W$
where $h$ is Planck's constant,$v$ is the frequency of incident radiation,and $W$ is the work function of the metal.
This equation represents a straight line $y = mx + c$ where the slope $m = h$ is constant for all metals.
Therefore,the graphs for different metals must be parallel straight lines with the same slope.
Given the work functions are in the order $W_A > W_B > W_C$,the threshold frequencies $v_0$ (where $E_k = 0$) follow the same order since $W = hv_0$:
$(v_0)_A > (v_0)_B > (v_0)_C$
This means the x-intercept for metal $A$ is the largest,followed by $B$,and then $C$.
Among the given options,the graph where the lines are parallel and the x-intercepts follow the order $A > B > C$ is the correct one. Based on the provided images,the graph that shows parallel lines with the x-intercepts in the order $A > B > C$ is represented by option $B$.

(C) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \phi + eV_s$,where $\phi = \frac{hc}{\lambda_{th}}$ is the work function and $V_s$ is the stopping potential.
For the first case: $\frac{hc}{\lambda} = \frac{hc}{\lambda_{th}} + e(3V_0)$ --- $(1)$
For the second case: $\frac{hc}{2\lambda} = \frac{hc}{\lambda_{th}} + eV_0$ --- $(2)$
Multiply equation $(2)$ by $3$: $\frac{3hc}{2\lambda} = \frac{3hc}{\lambda_{th}} + 3eV_0$ --- $(3)$
Subtract equation $(1)$ from equation $(3)$:
$\frac{3hc}{2\lambda} - \frac{hc}{\lambda} = \frac{3hc}{\lambda_{th}} - \frac{hc}{\lambda_{th}}$
$\frac{hc}{2\lambda} = \frac{2hc}{\lambda_{th}}$
$\frac{1}{2\lambda} = \frac{2}{\lambda_{th}}$
$\lambda_{th} = 4\lambda$.

(C) The energy of the incident photon is given by $E = h\nu$.
Given: $h = 6.63 \times 10^{-34} \, Js$,$\nu = 6 \times 10^{14} \, Hz$.
$E = (6.63 \times 10^{-34}) \times (6 \times 10^{14}) \, J = 39.78 \times 10^{-20} \, J$.
To convert this energy into $eV$,we divide by $1.6 \times 10^{-19} \, J/eV$:
$E = \frac{39.78 \times 10^{-20}}{1.6 \times 10^{-19}} \, eV = 2.48625 \, eV \approx 2.49 \, eV$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max} = E - \Phi$,where $\Phi$ is the work function.
Given $\Phi = 2 \, eV$.
$K_{max} = 2.49 \, eV - 2 \, eV = 0.49 \, eV$.

(B) The work function $\phi_0$ of a photoelectric emitter is related to the threshold wavelength $\lambda_0$ by the equation $\phi_0 = \frac{hc}{\lambda_0}$.
Given that the photoelectrons are just liberated,the incident wavelength is equal to the threshold wavelength.
For the first emitter,$\lambda_{01} = 300 \ nm$.
For the second emitter,$\lambda_{02} = 600 \ nm$.
The ratio of the work functions is $\frac{\phi_{01}}{\phi_{02}} = \frac{hc / \lambda_{01}}{hc / \lambda_{02}} = \frac{\lambda_{02}}{\lambda_{01}}$.
Substituting the values: $\frac{\phi_{01}}{\phi_{02}} = \frac{600 \ nm}{300 \ nm} = \frac{2}{1}$.
Thus,the ratio is $2 : 1$.

(B) The energy of the incident photon is given by $E = \frac{12400}{\lambda (\text{in } Å)} \, eV$.
Given $\lambda = 1240 \, Å$, so $E = \frac{12400}{1240} = 10 \, eV$.
According to Einstein's photoelectric equation, $E = W + K_{max}$, where $K_{max} = e V_s$.
Given stopping potential $V_s = 8 \, V$, so $K_{max} = 8 \, eV$.
Therefore, the work function $W = E - K_{max} = 10 \, eV - 8 \, eV = 2 \, eV$.
The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \frac{12400}{W (\text{in } eV)} \, Å$.
$\lambda_0 = \frac{12400}{2} = 6200 \, Å$.

(A) According to Einstein's photoelectric equation,the energy of the incident photon is given by $E = h\nu$,where $\nu$ is the frequency of the incident light.
The photoelectric equation is $E = \phi_0 + K_{\max}$,where $\phi_0 = h\nu_0$ is the work function and $K_{\max}$ is the maximum kinetic energy of the emitted photoelectrons.
Given,the frequency of incident light $\nu = 4\nu_0$ and the threshold frequency is $\nu_0$.
Substituting these values into the equation:
$h(4\nu_0) = h\nu_0 + K_{\max}$
Solving for $K_{\max}$:
$K_{\max} = 4h\nu_0 - h\nu_0$
$K_{\max} = 3h\nu_0$

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $KE_{\max}$ is given by $KE_{\max} = h n - \phi$,where $h$ is Planck's constant,$n$ is the frequency,and $\phi$ is the work function of the metal.
For the first case: $\frac{1}{2} m v^2 = h n - \phi$ ..... $(i)$
For the second case with frequency $3n$: $\frac{1}{2} m (v^{\prime})^2 = 3 h n - \phi$ ..... $(ii)$
From equation $(i)$,we have $h n = \frac{1}{2} m v^2 + \phi$. Substituting this into equation $(ii)$:
$\frac{1}{2} m (v^{\prime})^2 = 3 (\frac{1}{2} m v^2 + \phi) - \phi$
$\frac{1}{2} m (v^{\prime})^2 = \frac{3}{2} m v^2 + 2 \phi$
$(v^{\prime})^2 = 3 v^2 + \frac{4 \phi}{m}$
Since $\phi > 0$,it follows that $(v^{\prime})^2 > 3 v^2$,which implies $v^{\prime} > \sqrt{3} v$.
Therefore,the maximum velocity will be more than $\sqrt{3} v$.

(D) From Einstein's photoelectric equation,the energy of the incident photon is equal to the work function plus the maximum kinetic energy,which is $eV_s$ where $V_s$ is the stopping potential.
For wavelength $\lambda_1$: $\frac{hc}{\lambda_1} = \phi + eV$ ..... $(1)$
For wavelength $\lambda_2$: $\frac{hc}{\lambda_2} = \phi + 3eV$ ..... $(2)$
For wavelength $\lambda_3$: $\frac{hc}{\lambda_3} = \phi + eV'$ ..... $(3)$
Subtracting $(1)$ from $(2)$ to eliminate $\phi$:
$\frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} = 2eV \implies eV = \frac{hc}{2} \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right)$
Now,find $\phi$ from $(1)$:
$\phi = \frac{hc}{\lambda_1} - eV = \frac{hc}{\lambda_1} - \frac{hc}{2} \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right) = hc \left( \frac{1}{\lambda_1} - \frac{1}{2\lambda_2} + \frac{1}{2\lambda_1} \right) = hc \left( \frac{3}{2\lambda_1} - \frac{1}{2\lambda_2} \right)$
Substitute $\phi$ into $(3)$:
$eV' = \frac{hc}{\lambda_3} - \phi = \frac{hc}{\lambda_3} - hc \left( \frac{3}{2\lambda_1} - \frac{1}{2\lambda_2} \right)$
$V' = \frac{hc}{e} \left[ \frac{1}{\lambda_3} + \frac{1}{2\lambda_2} - \frac{3}{2\lambda_1} \right]$

(B) Given: $\lambda_1 = 4972\,\mathring{A}$,$\lambda_2 = 6216\,\mathring{A}$,Total Intensity $I = 3.6 \times 10^{-3}\,\text{W/m}^2$,Area $A = 1\,\text{cm}^2 = 10^{-4}\,\text{m}^2$,Work function $\phi = 2.3\,\text{eV}$.
Intensity per wavelength $I' = I/2 = 1.8 \times 10^{-3}\,\text{W/m}^2$.
Energy of photons for each wavelength:
$E_1 = \frac{hc}{\lambda_1} = \frac{12400}{4972} \approx 2.49\,\text{eV} > 2.3\,\text{eV}$ (Capable of emission).
$E_2 = \frac{hc}{\lambda_2} = \frac{12400}{6216} \approx 1.99\,\text{eV} < 2.3\,\text{eV}$ (Not capable of emission).
Only photons of $\lambda_1$ contribute to photoelectric emission.
Power incident from $\lambda_1$ is $P = I' \times A = 1.8 \times 10^{-3} \times 10^{-4} = 1.8 \times 10^{-7}\,\text{W}$.
Number of photons per second $n = \frac{P}{E_1} = \frac{1.8 \times 10^{-7}}{2.49 \times 1.6 \times 10^{-19}} \approx 4.5 \times 10^{11}\,\text{photons/s}$.
Since each capable photon ejects one electron,the number of photoelectrons liberated in $2\,\text{s}$ is $N = n \times 2 = 4.5 \times 10^{11} \times 2 = 9 \times 10^{11}$.

(A) According to Einstein's photoelectric equation: $eV_0 = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
Rearranging this,we get: $V_0 = \frac{hc}{e} \left(\frac{1}{\lambda}\right) - \frac{\phi}{e}$.
This is the equation of a straight line $y = mx + c$,where the slope $m = \frac{hc}{e}$ is constant for both metals,and the x-intercept is $\frac{1}{\lambda_0} = \frac{\phi}{hc}$.
From the graph,the x-intercept for metal $A$ is smaller than the x-intercept for metal $B$. Since the x-intercept is proportional to the work function $\phi$ (i.e.,$\phi = hc \cdot (1/\lambda_0)$),a smaller x-intercept implies a smaller work function.
Therefore,$\phi_A < \phi_B$,which means the work function of metal $B$ is greater than that of metal $A$.

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda} = \frac{12400}{\lambda(\text{in } \mathring{A})} \, eV$.
Substituting $\lambda = 2500\, \mathring{A}$,we get $E = \frac{12400}{2500} = 4.96\, eV$.
The maximum kinetic energy of the emitted electrons is $K_{max} = E - \Phi$,where $\Phi = 4.47\, eV$ is the work function.
$K_{max} = 4.96 - 4.47 = 0.49\, eV$.
As the ball emits electrons,it becomes positively charged,creating a potential $V$ that stops further emission. The stopping potential $V$ is equal to $K_{max}/e$,so $V = 0.49\, V$.
The potential of a charged sphere is $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} = k \frac{Q}{r}$,where $k = 9 \times 10^9\, N\cdot m^2/C^2$ and $r = 0.01\, m$.
$0.49 = \frac{9 \times 10^9 \times Q}{0.01}$.
$Q = \frac{0.49 \times 0.01}{9 \times 10^9} = \frac{0.0049}{9 \times 10^9} \approx 5.44 \times 10^{-13}\, C$.
Rounding to the nearest option,$Q = 5.5 \times 10^{-13}\, C$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = hv - hv_0$,where $h$ is Planck's constant,$v$ is the frequency of incident light,and $v_0$ is the threshold frequency.
If the frequency $v$ is doubled to $2v$,the new kinetic energy $K'_{max} = h(2v) - hv_0 = 2hv - hv_0$. This is not equal to $2K_{max} = 2(hv - hv_0) = 2hv - 2hv_0$. Thus,the maximum kinetic energy does not double.
Furthermore,the photocurrent depends on the number of photons incident per unit time,which is proportional to the intensity of the light,not the frequency. Therefore,Statement $1$ is false.
Statement $2$ correctly states that $K_{max}$ depends linearly on frequency and photocurrent depends on intensity. Thus,Statement $2$ is true.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E_{max})$ of the ejected photoelectrons is given by: $K.E_{max} = hv - W$,where $h$ is Planck's constant,$v$ is the frequency of incident light,and $W$ is the work function of the metal.
Option $A$ is correct because if $v < W/h$,then $hv < W$,meaning the incident energy is insufficient to overcome the work function.
Option $B$ is correct as the photoelectric effect is an instantaneous process.
Option $D$ is correct because $K.E_{max}$ depends only on the frequency of the incident light,not its intensity.
Option $C$ is incorrect because the maximum kinetic energy is $hv - W$,not $hv$.

(A) Let the work function be $\phi$. The energy of the incident photons is given by $E = \frac{1240}{\lambda} \ eV$.
For $\lambda_1 = 350 \ nm$,$E_1 = \frac{1240}{350} \approx 3.543 \ eV$.
For $\lambda_2 = 540 \ nm$,$E_2 = \frac{1240}{540} \approx 2.296 \ eV$.
According to Einstein's photoelectric equation,$KE_{max} = E - \phi$.
Let $v_1$ and $v_2$ be the maximum speeds. Given $v_1 = 2v_2$,so $KE_1 = 4 KE_2$.
$3.543 - \phi = 4(2.296 - \phi)$.
$3.543 - \phi = 9.184 - 4\phi$.
$3\phi = 9.184 - 3.543 = 5.641$.
$\phi = \frac{5.641}{3} \approx 1.88 \ eV$.
The closest value is $1.8 \ eV$.

(D) The magnetic field is given by $B = B_0 [\sin(\omega_1 t) + \sin(\omega_2 t)]$,where $\omega_1 = 3.14 \times 10^7 c$ and $\omega_2 = 6.28 \times 10^7 c$.
Since $\omega = 2\pi \nu$,the frequencies are $\nu_1 = \frac{3.14 \times 10^7 c}{2\pi} = 0.5 \times 10^7 c$ and $\nu_2 = \frac{6.28 \times 10^7 c}{2\pi} = 1.0 \times 10^7 c$.
The maximum frequency is $\nu_{\max} = 1.0 \times 10^7 \times (3 \times 10^8) = 3 \times 10^{15} \ Hz$.
The energy of the photon is $E = h\nu_{\max} = (6.63 \times 10^{-34} \ J\cdot s) \times (3 \times 10^{15} \ Hz) = 1.989 \times 10^{-18} \ J$.
Converting to $eV$: $E = \frac{1.989 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 12.43 \ eV$.
The maximum kinetic energy is $KE_{\max} = E - \phi = 12.43 \ eV - 4.7 \ eV = 7.73 \ eV$. The closest option is $7.72 \ eV$.

(C) $1$. Calculate the number of incident photons per second $(N_i)$:
Intensity $I = 16 \, mW/m^2 = 16 \times 10^{-3} \, W/m^2$.
Area $A = 1 \times 10^{-4} \, m^2$.
Energy of one photon $E = 10 \, eV = 10 \times 1.6 \times 10^{-19} \, J = 1.6 \times 10^{-18} \, J$.
Total power incident $P = I \times A = 16 \times 10^{-3} \times 10^{-4} = 16 \times 10^{-7} \, W$.
$N_i = P / E = (16 \times 10^{-7}) / (1.6 \times 10^{-18}) = 10^{12} \, \text{photons/second}$.
$2$. Calculate the number of emitted photoelectrons per second $(N_e)$:
$N_e = 10\% \text{ of } N_i = 0.1 \times 10^{12} = 10^{11} \, \text{electrons/second}$.
$3$. Calculate the maximum kinetic energy $(K_{\max})$:
Using Einstein's photoelectric equation: $K_{\max} = E - \phi = 10 \, eV - 5 \, eV = 5 \, eV$.

(C) According to Einstein's photoelectric equation:
$\frac{hc}{\lambda_{1}} = \phi + eV_{1}$ ...... $(i)$
$\frac{hc}{\lambda_{2}} = \phi + eV_{2}$ ...... $(ii)$
Subtracting equation $(ii)$ from $(i)$:
$\frac{hc}{\lambda_{1}} - \frac{hc}{\lambda_{2}} = e(V_{1} - V_{2})$
$hc \left( \frac{1}{\lambda_{1}} - \frac{1}{\lambda_{2}} \right) = e \Delta V$
$\Delta V = \frac{hc}{e} \left( \frac{\lambda_{2} - \lambda_{1}}{\lambda_{1} \lambda_{2}} \right)$
Given $\frac{hc}{e} = 1240\, nm \cdot V$,$\lambda_{1} = 300\, nm$,and $\lambda_{2} = 400\, nm$:
$\Delta V = 1240 \times \left( \frac{400 - 300}{300 \times 400} \right)$
$\Delta V = 1240 \times \left( \frac{100}{120000} \right)$
$\Delta V = \frac{1240}{1200} \approx 1.03\, V$
Thus,the decrease in stopping potential is close to $1\, V$.

(D) According to Einstein's photoelectric equation,the stopping potential $V_s$ is related to the frequency $v$ by the equation: $eV_s = hv - \phi$,where $\phi = hv_0$ is the work function and $v_0$ is the threshold frequency.
For the first case: $e(V_0/2) = hv - \phi$ ..... $(1)$
For the second case: $e(V_0) = h(v/2) - \phi$ ..... $(2)$
Note: The stopping potential is given as negative,which indicates the magnitude of the potential required to stop the electrons. Thus,$eV_s = h(v - v_0)$.
From $(1)$: $eV_0/2 = hv - hv_0 \Rightarrow eV_0 = 2hv - 2hv_0$
From $(2)$: $eV_0 = hv/2 - hv_0$
Equating the two expressions for $eV_0$:
$2hv - 2hv_0 = hv/2 - hv_0$
$2hv - hv/2 = 2hv_0 - hv_0$
$3hv/2 = hv_0$
$v_0 = 3v/2$.

(A) The given electric field equation is in the form $\vec E = E_0 \cos(kx - \omega t)$.
Comparing the given equation,the angular frequency is $\omega = 2\pi \times 6 \times 10^{14} \, rad/s$.
The frequency $f$ is given by $f = \frac{\omega}{2\pi} = 6 \times 10^{14} \, Hz$.
The energy of a photon is $E = hf = \frac{hc}{\lambda}$. Using $E(eV) \approx \frac{12400}{\lambda(\mathring{A})}$,we first find $\lambda = \frac{c}{f} = \frac{3 \times 10^8}{6 \times 10^{14}} = 5 \times 10^{-7} \, m = 5000 \, \mathring{A}$.
Thus,$E = \frac{12400}{5000} = 2.48 \, eV$.
Using Einstein's photoelectric equation: $K_{max} = E - \phi$,where $\phi = 2 \, eV$.
$eV_s = 2.48 - 2 = 0.48 \, eV$.
Therefore,the stopping potential $V_s = 0.48 \, V$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:
$K_{\max} = E - \phi$
where $E$ is the energy of the incident photon and $\phi$ is the work function.
Given $E = \frac{1237}{\lambda}$ and $\phi = \frac{1237}{\lambda_0}$,where $\lambda_0 = 380 \, nm$ and $\lambda = 260 \, nm$.
$K_{\max} = \frac{1237}{260} - \frac{1237}{380}$
$K_{\max} = 1237 \times \left( \frac{380 - 260}{380 \times 260} \right)$
$K_{\max} = 1237 \times \left( \frac{120}{98800} \right)$
$K_{\max} \approx 1.5 \, eV$.

(B) According to Einstein's photoelectric equation:
$h\nu = \phi + eV_0$
Rearranging for the stopping potential $V_0$:
$V_0 = \frac{h\nu}{e} - \frac{\phi}{e}$
From the given graph,the threshold frequency $(\nu_0)$ is the frequency at which the stopping potential $V_0$ becomes zero. Looking at the graph,$V_0 = 0$ when $\nu = 4 \times 10^{14} \, Hz$.
At this point:
$0 = \frac{h\nu_0}{e} - \frac{\phi}{e}$
$\Rightarrow \phi = h\nu_0$
Substituting the values:
$\phi = (6.63 \times 10^{-34} \, Js) \times (4 \times 10^{14} \, Hz)$
$\phi = 26.52 \times 10^{-20} \, J$
To convert the work function into electron-volts $(eV)$:
$\phi (eV) = \frac{26.52 \times 10^{-20} \, J}{1.6 \times 10^{-19} \, C}$
$\phi = 1.6575 \, eV \approx 1.66 \, eV$

(A) According to Einstein's photoelectric equation,$hf = \phi_0 + KE_{max}$.
For the $1^{\text{st}}$ case: $hf = \phi_0 + K_0$ --- $(1)$
For the $2^{\text{nd}}$ case,the frequency is increased by a factor of $n_1$,so the new frequency is $n_1f$. The new $KE_{max}$ is $n_2K_0$:
$n_1hf = \phi_0 + n_2K_0$ --- $(2)$
Substitute $hf$ from equation $(1)$ into equation $(2)$:
$n_1(\phi_0 + K_0) = \phi_0 + n_2K_0$
Expand the equation:
$n_1\phi_0 + n_1K_0 = \phi_0 + n_2K_0$
Rearrange to solve for $\phi_0$:
$n_1\phi_0 - \phi_0 = n_2K_0 - n_1K_0$
$\phi_0(n_1 - 1) = K_0(n_2 - n_1)$
Therefore,the work function is:
$\phi_0 = \left( \frac{n_2 - n_1}{n_1 - 1} \right) K_0$

(B) From Einstein's photoelectric equation:
$4.25 = \phi_{A} + T_{A}$ --- $(1)$
$4.70 = \phi_{B} + T_{B}$ --- $(2)$
Given: $T_{B} = T_{A} - 1.50 \, eV$
De-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.
Thus,$\frac{\lambda_{A}}{\lambda_{B}} = \sqrt{\frac{T_{B}}{T_{A}}}$.
Given $\lambda_{B} = 2\lambda_{A}$,so $\frac{1}{2} = \sqrt{\frac{T_{B}}{T_{A}}} \Rightarrow \frac{1}{4} = \frac{T_{B}}{T_{A}} \Rightarrow T_{A} = 4T_{B}$.
Substituting $T_{A} = 4T_{B}$ into the given relation: $T_{B} = 4T_{B} - 1.50 \Rightarrow 3T_{B} = 1.50 \Rightarrow T_{B} = 0.50 \, eV$.
Then $T_{A} = 4 \times 0.50 = 2.0 \, eV$.
From $(1)$,$\phi_{A} = 4.25 - 2.0 = 2.25 \, eV$.
From $(2)$,$\phi_{B} = 4.70 - 0.50 = 4.20 \, eV$.
Therefore,the work function of $B$ is $4.20 \, eV$.

(B) According to Einstein's photoelectric equation,the kinetic energy of emitted electrons 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.
Since $K_{max} = eV_s$,where $V_s$ is the stopping potential and $e$ is the charge of an electron,we can write:
$eV_s = h\nu - \phi$
$V_s = (\frac{h}{e})\nu - \frac{\phi}{e}$
Comparing this with the equation of a straight line $y = mx + c$,the slope $m$ is equal to $\frac{h}{e}$.
Given that the slope is $4.12 \times 10^{-15} \, V-s$ and the charge of an electron $e = 1.6 \times 10^{-19} \, C$,we have:
$\frac{h}{e} = 4.12 \times 10^{-15}$
$h = (4.12 \times 10^{-15}) \times (1.6 \times 10^{-19})$
$h \approx 6.592 \times 10^{-34} \, J-s \approx 6.6 \times 10^{-34} \, J-s$.

(C) The photoelectric effect occurs when the frequency of incident light is greater than the threshold frequency, or equivalently, when the wavelength is less than the threshold wavelength $ (\lambda < \lambda_{th}) $.
Given that green light causes emission but yellow light does not, the threshold wavelength $ (\lambda_{th}) $ lies between the wavelengths of green and yellow light.
In the visible spectrum, the order of wavelengths is: $ \lambda_{Red} > \lambda_{Orange} > \lambda_{Yellow} > \lambda_{Green} > \lambda_{Blue} > \lambda_{Indigo} > \lambda_{Violet} $.
Since the wavelength of indigo light is smaller than that of green light, it has a higher frequency and energy, thus it will cause the emission of photo-electrons.

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $hc = 12420 \, eV \cdot \mathring{A}$ (or $12375 \, eV \cdot \mathring{A}$ as per given constants),we calculate $E = \frac{12375}{2000} \, eV = 6.1875 \, eV$.
According to Einstein's photoelectric equation,$E = W_0 + K_{\max}$,where $W_0$ is the work function and $K_{\max}$ is the maximum kinetic energy.
$K_{\max} = E - W_0 = 6.1875 \, eV - 5.01 \, eV = 1.1775 \, eV$.
The stopping potential $V_s$ is related to the maximum kinetic energy by $K_{\max} = e V_s$.
Therefore,$V_s = 1.1775 \, V$,which is approximately $1.2 \, V$.

(B) According to Einstein's photoelectric equation, the maximum kinetic energy $K$ is given by $K = h v - \Phi$, where $\Phi = h v_0$ is the work function and $v_0$ is the threshold frequency.
For frequency $v_1$, $K_1 = h v_1 - h v_0 = h(v_1 - v_0)$.
For frequency $v_2$, $K_2 = h v_2 - h v_0 = h(v_2 - v_0)$.
Given the ratio of kinetic energies is $K_1 / K_2 = 1 / x$, we have $x K_1 = K_2$.
Substituting the expressions: $x h(v_1 - v_0) = h(v_2 - v_0)$.
$x v_1 - x v_0 = v_2 - v_0$.
$x v_1 - v_2 = x v_0 - v_0$.
$x v_1 - v_2 = v_0(x - 1)$.
Therefore, the threshold frequency $v_0 = \frac{x v_1 - v_2}{x - 1}$.

(C) The energy of the incident photon is given by $E = \frac{12375}{\lambda (\text{in } Å)} \, eV$.
Substituting the given wavelength, $E = \frac{12375}{5000} = 2.475 \, eV$.
According to Einstein's photoelectric equation, $E = W_0 + K_{\max}$, where $W_0$ is the work function and $K_{\max}$ is the maximum kinetic energy of the photoelectrons.
The maximum kinetic energy is related to the stopping potential $V_0$ by $K_{\max} = eV_0$.
Given $V_0 = 1.36 \, V$, we have $K_{\max} = 1.36 \, eV$.
Substituting the values into the equation: $2.475 = W_0 + 1.36$.
Solving for $W_0$: $W_0 = 2.475 - 1.36 = 1.115 \, eV \approx 1.12 \, eV$.

(C) According to Einstein's photoelectric equation,the energy of the incident photon is equal to the sum of the work function and the maximum kinetic energy of the emitted electron:
$E = \phi + K_{max}$
Substituting $E = \frac{hc}{\lambda}$ and $K_{max} = \frac{1}{2}mv^2$,we get:
$\frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2$
Rearranging the terms to solve for $v$:
$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi$
$\frac{1}{2}mv^2 = \frac{hc - \lambda \phi}{\lambda}$
$v^2 = \frac{2(hc - \lambda \phi)}{m \lambda}$
$v = [\frac{2(hc - \lambda \phi)}{m \lambda}]^{1/2}$

(B) In the photoelectric effect,electrons are emitted from the emitter plate and move towards the collector plate.
When an electric field is applied in the vertically downward direction,the electrons (being negatively charged) experience an electrostatic force in the upward direction (opposite to the electric field).
Since the collector plate is placed vertically above the emitter plate,this upward force acts in the direction of the motion of the electrons.
As a result,the electrons are accelerated,which increases their kinetic energy upon reaching the collector plate.
Because the kinetic energy of the emitted electrons increases,a higher retarding potential (stopping potential) is required to stop them.
Therefore,the stopping potential increases,while the saturation photocurrent and the threshold wavelength remain unchanged.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = e V_0$,where $V_0$ is the stopping potential.
Einstein's equation is $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
Given: $\phi = 2.14\, eV$,$V_0 = 0.60\, V$,so $K_{max} = 0.60\, eV$.
Substituting the values: $0.60\, eV = \frac{1240\, eV \cdot nm}{\lambda} - 2.14\, eV$.
$0.60 + 2.14 = \frac{1240}{\lambda}$.
$2.74 = \frac{1240}{\lambda}$.
$\lambda = \frac{1240}{2.74} \approx 452.55\, nm$.
Since $452.55\, nm$ is not among the options,the correct choice is $D$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = h\nu - \Phi$,where $\nu$ is the frequency and $\Phi$ is the work function.
Initially,$K_0 = h\nu - \Phi$ --- $(1)$
When the frequency is increased by a factor of $n_1$,the new frequency is $\nu' = n_1\nu$. The new maximum kinetic energy is $n_2K_0$.
So,$n_2K_0 = h(n_1\nu) - \Phi$ --- $(2)$
From $(1)$,$h\nu = K_0 + \Phi$. Substituting this into $(2)$:
$n_2K_0 = n_1(K_0 + \Phi) - \Phi$
$n_2K_0 = n_1K_0 + n_1\Phi - \Phi$
$n_2K_0 - n_1K_0 = \Phi(n_1 - 1)$
$K_0(n_2 - n_1) = \Phi(n_1 - 1)$
$\Phi = \left( \frac{n_2 - n_1}{n_1 - 1} \right) K_0$
Thus,the correct option is $A$.

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda} = \frac{1240 \, eV \cdot nm}{400 \, nm} = 3.1 \, eV$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is $K_{max} = E - \phi$,where $\phi = 2.5 \, eV$.
$K_{max} = 3.1 \, eV - 2.5 \, eV = 0.6 \, eV$.
Converting $K_{max}$ to Joules: $K_{max} = 0.6 \times 1.6 \times 10^{-19} \, J = 0.96 \times 10^{-19} \, J$.
The relationship between momentum $p$ and kinetic energy $K$ is $p = \sqrt{2mK}$.
$p = \sqrt{2 \times (9.1 \times 10^{-31} \, kg) \times (0.96 \times 10^{-19} \, J)}$.
$p = \sqrt{17.472 \times 10^{-50}} \approx 4.18 \times 10^{-25} \, kg \cdot m/s$.
Rounding to the nearest option,the value is approximately $4 \times 10^{-25} \, kg \cdot m/s$.

(B) The energy of the incident photon is given by $E = hf$.
The work function of the emitter is given by $\phi = hf_0$,where $f_0$ is the threshold frequency.
According to Einstein's photoelectric equation,the energy of the incident photon is equal to the sum of the work function and the maximum kinetic energy $(K_{max})$ of the emitted electron:
$hf = \phi + K_{max}$
Substituting the value of the work function:
$hf = hf_0 + K_{max}$
Rearranging the equation to solve for $K_{max}$:
$K_{max} = hf - hf_0$
$K_{max} = h(f - f_0)$

(B) According to Einstein's photoelectric equation, $K_{max} = h\nu - \Phi$, where $\nu$ is the frequency of incident light and $\Phi$ is the work function of the metal.
Case $1$: $K_0 = h\nu - \Phi$ --- $(1)$
Case $2$: $3K_0 = h(2\nu) - \Phi = 2h\nu - \Phi$ --- $(2)$
Subtracting equation $(1)$ from $(2)$:
$(3K_0 - K_0) = (2h\nu - \Phi) - (h\nu - \Phi)$
$2K_0 = h\nu$
Substituting $h\nu = 2K_0$ into equation $(1)$:
$K_0 = 2K_0 - \Phi$ implies $\Phi = K_0$
Case $3$: If frequency is tripled, $\nu' = 3\nu$.
$K_{max}' = h(3\nu) - \Phi = 3(h\nu) - \Phi$
Substituting $h\nu = 2K_0$ and $\Phi = K_0$:
$K_{max}' = 3(2K_0) - K_0 = 6K_0 - K_0 = 5K_0$.
Therefore, the value is $5$.

(B) 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 and $\phi_0$ is the work function of the metal.
Given that the threshold frequency is $v$,the work function is $\phi_0 = hv$.
The energy of the incident light with frequency $4v$ is $E = h(4v) = 4hv$.
Substituting these values into the equation:
$K_{\max} = 4hv - hv = 3hv$.

(A) According to Einstein's photoelectric equation,when a photon of frequency $v$ strikes a metal surface,the energy of the photon is used to overcome the work function $\phi$ and the remainder is given as the maximum kinetic energy $(KE)_{\max}$ to the emitted photoelectron.
The energy of the incident photon is given by $E = hv$,where $h$ is Planck's constant.
The work function of the metal is given by $\phi = hv_0$,where $v_0$ is the threshold frequency.
Applying the law of conservation of energy:
$hv = (KE)_{\max} + \phi$
Substituting the value of $\phi$:
$hv = (KE)_{\max} + hv_0$
Rearranging the equation for maximum kinetic energy:
$(KE)_{\max} = hv - hv_0$
$(KE)_{\max} = h(v - v_0)$
Since $h$ is a constant,the maximum kinetic energy is directly proportional to $(v - v_0)$.