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

(B) 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.
For frequency $\nu_1$,$K_1 = h(\nu_1 - \nu_0) \dots (i)$
For frequency $\nu_2$,$K_2 = h(\nu_2 - \nu_0) \dots (ii)$
Given the ratio $\frac{K_1}{K_2} = \frac{1}{k}$,we have:
$\frac{h(\nu_1 - \nu_0)}{h(\nu_2 - \nu_0)} = \frac{1}{k}$
$k(\nu_1 - \nu_0) = \nu_2 - \nu_0$
$k\nu_1 - k\nu_0 = \nu_2 - \nu_0$
$k\nu_1 - \nu_2 = k\nu_0 - \nu_0$
$k\nu_1 - \nu_2 = \nu_0(k - 1)$
$\nu_0 = \frac{k\nu_1 - \nu_2}{k - 1}$

(C) The kinetic energy of the fastest photoelectron is equal to the work done by the stopping potential $V_s$:
$\frac{1}{2} m v_{\max}^2 = e V_s$
Rearranging for the stopping potential $V_s$:
$V_s = \frac{m v_{\max}^2}{2e} = \frac{v_{\max}^2}{2(e/m)}$
Given $v_{\max} = 1.2 \times 10^6 \ m/s$ and specific charge $(e/m) = 1.8 \times 10^{11} \ C/kg$:
$V_s = \frac{(1.2 \times 10^6)^2}{2 \times 1.8 \times 10^{11}}$
$V_s = \frac{1.44 \times 10^{12}}{3.6 \times 10^{11}}$
$V_s = \frac{14.4}{3.6} = 4 \ V$

(D) Given: Work function $\Phi = 1 \ eV$,wavelength $\lambda = 3000 \ \mathring{A}$,and specific charge of electron $\frac{e}{m} = 1.76 \times 10^{11} \ C/kg$.
Using Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \Phi$.
Energy of incident photon $E = \frac{12400}{\lambda(\text{in } \mathring{A})} \ eV \approx \frac{12300}{3000} \ eV = 4.1 \ eV$.
Maximum kinetic energy $K_{max} = 4.1 \ eV - 1 \ eV = 3.1 \ eV$.
Converting to Joules: $K_{max} = 3.1 \times 1.6 \times 10^{-19} \ J$.
Using $K_{max} = \frac{1}{2}mv^2$,we get $v = \sqrt{\frac{2 K_{max}}{m}} = \sqrt{2 \times K_{max} \times \frac{e}{m} \times \frac{1}{e}}$.
$v = \sqrt{2 \times 3.1 \times 1.76 \times 10^{11}} = \sqrt{10.912 \times 10^{11}} \approx \sqrt{1.0912} \times 10^6 \ m/s \approx 1.04 \times 10^6 \ m/s$.
The closest option is $10^6 \ m/s$.

(B) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
Given for $\lambda_1 = 10,000\;\mathring A$,$E_1 = 1.23\; eV$.
For $\lambda_2 = 5000\;\mathring A$,the energy $E_2$ is:
$E_2 = E_1 \times \frac{\lambda_1}{\lambda_2} = 1.23 \times \frac{10,000}{5,000} = 1.23 \times 2 = 2.46\; eV$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max} = eV_s = E_2 - \phi$,where $\phi$ is the work function.
Given stopping potential $V_s = 1.36\; V$,so $K_{max} = 1.36\; eV$.
Substituting the values: $1.36 = 2.46 - \phi$.
Therefore,$\phi = 2.46 - 1.36 = 1.10\; eV$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = hf - hf_0 = h(f - f_0)$.
Given frequencies are $f_1 = 5.5 \times 10^8 \text{ MHz}$ and $f_2 = 4.5 \times 10^8 \text{ MHz}$.
The ratio of maximum kinetic energies is $\frac{K_1}{K_2} = \frac{1}{5}$.
Substituting the values: $\frac{K_1}{K_2} = \frac{h(f_1 - f_0)}{h(f_2 - f_0)} = \frac{f_1 - f_0}{f_2 - f_0} = \frac{1}{5}$.
$\frac{5.5 \times 10^8 - f_0}{4.5 \times 10^8 - f_0} = \frac{1}{5}$.
$5(5.5 \times 10^8 - f_0) = 4.5 \times 10^8 - f_0$.
$27.5 \times 10^8 - 5f_0 = 4.5 \times 10^8 - f_0$.
$4f_0 = 23.0 \times 10^8$.
$f_0 = 5.75 \times 10^8 \text{ MHz}$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy of an emitted photoelectron is given by $K_{max} = hv - W$,where $h$ is Planck's constant,$v$ is the frequency of incident light,and $W$ is the work function of the metal.
For photoemission to occur,the incident photon energy $hv$ must be greater than or equal to the work function $W$ $(hv \ge W)$.
If the frequency $v$ of the incident light is less than the threshold frequency $v_0 = W/h$,then $hv < W$,and no photoelectrons will be emitted from the metal surface.
Therefore,the condition for the emission of photoelectrons is that the frequency $v$ must be at least $W/h$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ of emitted photoelectrons is given by $K = hv - E_0$,where $hv$ is the energy of the incident photon and $E_0$ is the work function of the metal.
For the first case: $K = hv - E_0$ --- $(1)$
For the second case,the incident energy is $2hv$. Let the new maximum kinetic energy be $K'$.
$K' = 2hv - E_0$ --- $(2)$
From equation $(1)$,we have $E_0 = hv - K$.
Substituting this value of $E_0$ into equation $(2)$:
$K' = 2hv - (hv - K)$
$K' = 2hv - hv + K$
$K' = hv + K$

(C) The work function $\Phi_0$ is given as $4.0 \ eV$.
For photoelectric emission,the energy of the incident photon $E$ must be at least equal to the work function $\Phi_0$.
$E = \frac{hc}{\lambda} \geq \Phi_0$.
To find the longest wavelength $\lambda_{\max}$,we set $E = \Phi_0$.
Using the relation $\lambda (nm) = \frac{1240}{\Phi_0 (eV)}$:
$\lambda_{\max} = \frac{1240}{4.0} \ nm$.
$\lambda_{\max} = 310 \ nm$.

(B) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Substituting the values: $E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3500 \times 10^{-10}} \ J$.
Converting to electron-volts $(eV)$: $E = \frac{12400 \ \mathring A \cdot eV}{3500 \ \mathring A} \approx 3.54 \ eV$.
Photoelectric emission occurs if the incident energy $E$ is greater than the work function $\phi$ of the metal $(E > \phi)$.
For metal $A$: $\phi_A = 4.2 \ eV$. Since $3.54 \ eV < 4.2 \ eV$,metal $A$ will not emit photoelectrons.
For metal $B$: $\phi_B = 1.9 \ eV$. Since $3.54 \ eV > 1.9 \ eV$,metal $B$ will emit photoelectrons.
Therefore,only metal $B$ will emit photoelectrons.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K.E._{max} = h\nu - \phi$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\phi$ is the work function of the material.
$1$. The kinetic energy depends on the frequency of incident light $(\nu)$ or its wavelength $(\lambda = c/\nu)$.
$2$. The kinetic energy depends on the work function $(\phi)$,which is determined by the nature of the material.
$3$. The kinetic energy does not depend on the intensity of the incident light. The intensity only affects the number of photoelectrons emitted per unit time.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = h\nu - \Phi$,where $\Phi$ is the work function of the metal.
Since $\nu = c / \lambda$,reducing the wavelength $\lambda$ increases the frequency $\nu$ of the incident radiation.
An increase in frequency $\nu$ leads to an increase in the maximum kinetic energy $K_{max}$ of the photoelectrons.
Since the stopping potential $V_s$ is related to the maximum kinetic energy by $K_{max} = eV_s$,an increase in $K_{max}$ results in an increase in the stopping potential $V_s$.
Therefore,the stopping potential increases.

(C) The work function $\phi_0$ is given by the formula: $\phi_0 = h \nu_0 = \frac{hc}{\lambda_0}$.
Given: $h = 6.6 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $\lambda_0 = 5000 \ \mathring A = 5000 \times 10^{-10} \ m = 5 \times 10^{-7} \ m$.
Substituting the values:
$\phi_0 = \frac{(6.6 \times 10^{-34}) \times (3 \times 10^8)}{5 \times 10^{-7}} \ J$.
$\phi_0 = \frac{19.8 \times 10^{-26}}{5 \times 10^{-7}} \ J = 3.96 \times 10^{-19} \ J$.
To convert this into $eV$,divide by $1.6 \times 10^{-19} \ J/eV$:
$\phi_0 = \frac{3.96 \times 10^{-19}}{1.6 \times 10^{-19}} \ eV = 2.475 \ eV$.
Rounding to the nearest value,we get $\phi_0 \approx 2.5 \ eV$.

(C) Given: Maximum velocity $v_{max} = 4.8 \ m/s$,specific charge $e/m = 1.76 \times 10^{11} \ C/kg$.
The stopping potential $V_s$ is related to the maximum kinetic energy by the equation: $eV_s = \frac{1}{2}mv_{max}^2$.
Rearranging for $V_s$,we get: $V_s = \frac{1}{2} \left(\frac{m}{e}\right) v_{max}^2$.
Substituting the values: $V_s = \frac{1}{2} \times \left(\frac{1}{1.76 \times 10^{11}}\right) \times (4.8)^2$.
$V_s = \frac{23.04}{3.52 \times 10^{11}} \approx 6.54 \times 10^{-11} \ V$.
Rounding to the nearest significant value provided in the options,we get $V_s \approx 7 \times 10^{-11} \ V$.

(A) According to Einstein's photoelectric equation, $K_{max} = h\nu - \phi$, where $\phi$ is the work function. Since the frequency of $X$-rays is much higher than that of ultraviolet light, $K_{max}$ increases. Since $K_{max} = eV_0$, the stopping potential $V_0$ also increases. Thus, Statement-$1$ is true.
Statement-$2$ is false because photoelectrons are emitted with a range of speeds not because of a range of incident frequencies, but because electrons are emitted from different depths within the metal, losing varying amounts of energy due to collisions before escaping the surface. Even with monochromatic light, a range of speeds is observed.

(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $hc \approx 12400 \ eV \cdot \mathring A$,we get $E = \frac{12400}{6600} \ eV \approx 1.88 \ eV$.
The work function of the metal is $\Phi_0 = 2 \ eV$.
According to Einstein's photoelectric equation,$K_{max} = E - \Phi_0$.
Since the energy of the incident photon $(1.88 \ eV)$ is less than the work function $(2 \ eV)$,the incident light does not have enough energy to eject photoelectrons from the metal surface.
Therefore,no photoelectrons are emitted.

(A) The work function $\Phi$ is related to the threshold wavelength $\lambda_0$ by the formula $\Phi = \frac{hc}{\lambda_0}$.
For material $A$,$\Phi_A = 2.5 \ eV = \frac{hc}{\lambda}$.
For material $B$,$\Phi_B = 5 \ eV = \frac{hc}{\lambda_B}$.
Dividing the two equations: $\frac{\Phi_A}{\Phi_B} = \frac{hc/\lambda}{hc/\lambda_B} = \frac{\lambda_B}{\lambda}$.
Substituting the values: $\frac{2.5}{5} = \frac{\lambda_B}{\lambda}$.
$0.5 = \frac{\lambda_B}{\lambda} \Rightarrow \lambda_B = \frac{\lambda}{2}$.

(A) According to Einstein's photoelectric equation: $K_{max} = hf - \phi_0$.
Since $K_{max} = eV_0$, we have $eV_0 = hf - \phi_0$.
Dividing by $e$, we get $V_0 = (h/e)f - (\phi_0/e)$.
Comparing this with the equation of a straight line $y = mx + c$, where $y = V_0$ and $x = f$, the slope $m$ is equal to $h/e$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by:
$K_{max} = \frac{hc}{\lambda} - \phi_0$
Where $h$ is Planck's constant,$c$ is the speed of light,$\lambda$ is the wavelength,and $\phi_0$ is the work function.
Given $\lambda = 400 \ nm$ and $K_{max} = 1.68 \ eV$.
Using the relation $hc \approx 1240 \ eV \cdot nm$:
Energy of incident photon $E = \frac{1240 \ eV \cdot nm}{400 \ nm} = 3.10 \ eV$.
Now,substituting the values into the equation:
$\phi_0 = E - K_{max}$
$\phi_0 = 3.10 \ eV - 1.68 \ eV = 1.42 \ eV$.
Therefore,the work function of the metal is $1.42 \ eV$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{\max})$ of emitted photoelectrons is given by:
$K_{\max} = E - \phi_0$
Where $E = hf$ is the energy of the incident photon and $\phi_0 = hf_0$ is the work function of the surface.
Substituting these values:
$K_{\max} = hf - hf_0$
$K_{\max} = h(f - f_0)$

(A) According to Einstein's photoelectric equation,the stopping potential $(V_0)$ is given by the relation:
$V_0 = (\frac{h}{e})f - (\frac{h}{e})f_0$
where $h$ is Planck's constant,$e$ is the charge of an electron,$f$ is the frequency of incident radiation,and $f_0$ is the threshold frequency.
This equation is of the form $y = mx + c$,which represents a straight line.
Therefore,the graph of $V_0$ versus $f$ is a straight line.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi_0$.
Since $K_{max} = eV_0$,where $V_0$ is the stopping potential,we have $eV_0 = \frac{hc}{\lambda} - \phi_0$.
First,calculate the energy of the incident photon in $eV$:
$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{200 \times 10^{-9} \times 1.6 \times 10^{-19}} \ eV$.
$E = \frac{19.878 \times 10^{-26}}{320 \times 10^{-28}} \ eV = \frac{19.878}{3.2} \ eV \approx 6.21 \ eV$.
Now,calculate the stopping potential:
$eV_0 = 6.21 \ eV - 4.71 \ eV = 1.50 \ eV$.
Therefore,the stopping potential $V_0 = 1.50 \ V$,which corresponds to $1.5 \ eV$ in terms of energy units.

(D) 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.
Case $1$: $E_1 = 2hv_0$ and $\Phi = hv_0$.
$K_{max1} = 2hv_0 - hv_0 = hv_0 = \frac{1}{2}mv_1^2$,where $v_1 = 4 \times 10^6 \, m/s$.
Case $2$: $E_2 = 5hv_0$ and $\Phi = hv_0$.
$K_{max2} = 5hv_0 - hv_0 = 4hv_0 = \frac{1}{2}mv_2^2$.
Dividing the two equations:
$\frac{\frac{1}{2}mv_2^2}{\frac{1}{2}mv_1^2} = \frac{4hv_0}{hv_0} = 4$.
$\frac{v_2^2}{v_1^2} = 4 \Rightarrow \frac{v_2}{v_1} = 2$.
$v_2 = 2v_1 = 2 \times (4 \times 10^6 \, m/s) = 8 \times 10^6 \, m/s$.

(D) The photoelectric effect cannot be explained by the classical wave theory of light. It can only be explained by the quantum theory of light,which states that light consists of discrete packets of energy called photons. When a photon of sufficient energy strikes a metal surface,it transfers its energy to an electron,causing it to be emitted.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{\max})$ of emitted photoelectrons is related to the stopping potential $(V_0)$ by the equation:
$K_{\max} = e V_0$
Given that the maximum kinetic energy $K_{\max} = 4 \, eV$.
Substituting the values:
$4 \, eV = e V_0$
$V_0 = 4 \, V$
Therefore,the stopping potential is $4 \, V$.

(B) Let the work functions of metals $A$ and $B$ be $\phi_A$ and $\phi_B$ respectively. Given $\frac{\phi_A}{\phi_B} = \frac{1}{2}$,so $\phi_B = 2\phi_A$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = h\nu - \phi$.
For metal $A$: $K_A = hf - \phi_A$ ---$(1)$
For metal $B$: $K_B = h(2f) - \phi_B = 2hf - 2\phi_A$ ---$(2)$
From equation $(2)$,we can factor out $2$: $K_B = 2(hf - \phi_A)$.
Substituting equation $(1)$ into this,we get $K_B = 2K_A$.
Therefore,the ratio $\frac{K_A}{K_B} = \frac{1}{2}$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = E - \Phi$,where $E$ is the incident photon energy and $\Phi$ is the work function.
Given: $E_1 = 1 \ eV$,$E_2 = 2.5 \ eV$,and $\Phi = 0.5 \ eV$.
For the first photon: $K_{\max 1} = E_1 - \Phi = 1 \ eV - 0.5 \ eV = 0.5 \ eV$.
For the second photon: $K_{\max 2} = E_2 - \Phi = 2.5 \ eV - 0.5 \ eV = 2.0 \ eV$.
The ratio of the maximum kinetic energies is $\frac{K_{\max 1}}{K_{\max 2}} = \frac{0.5}{2.0} = \frac{1}{4}$.
Thus,the ratio is $1:4$.

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$. Using $hc \approx 12400 \, eV \cdot \mathring{A}$,we have $E = \frac{12400}{3300} \approx 3.76 \, eV$.
Photoelectric emission occurs if the incident photon energy $E$ is greater than the work function $\phi$ of the metal $(E > \phi)$.
Alternatively,the threshold wavelength is $\lambda_0 = \frac{hc}{\phi}$. Emission occurs if $\lambda < \lambda_0$.
Calculating $\lambda_0$ for each metal:
For $Na$: $\lambda_0 = \frac{12400}{1.92} \approx 6458 \, \mathring{A}$. Since $3300 < 6458$,emission occurs.
For $K$: $\lambda_0 = \frac{12400}{2.15} \approx 5767 \, \mathring{A}$. Since $3300 < 5767$,emission occurs.
For $Mo$: $\lambda_0 = \frac{12400}{4.17} \approx 2973 \, \mathring{A}$. Since $3300 > 2973$,no emission occurs.
For $Ni$: $\lambda_0 = \frac{12400}{5.0} = 2480 \, \mathring{A}$. Since $3300 > 2480$,no emission occurs.
Thus,$Mo$ and $Ni$ will not show photoelectric emission.

(C) The intensity of incident light is given by $I = \frac{E}{At} = \frac{nhc}{At\lambda}$,where $n$ is the number of photons incident per second.
The number of photons incident per second is $n = \frac{IA\lambda}{hc}$.
Given that $1 \%$ of incident photons emit photoelectrons,the number of photoelectrons emitted per second $N$ is:
$N = \frac{1}{100} \times n = \frac{1}{100} \times \frac{IA\lambda}{hc}$.
Substituting the given values:
$I = 1.0 \ W/m^2$,$A = 1 \ cm^2 = 10^{-4} \ m^2$,$\lambda = 300 \ nm = 300 \times 10^{-9} \ m$,$h = 6.63 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$.
$N = \frac{1}{100} \times \frac{1.0 \times 10^{-4} \times 300 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8}$.
$N = \frac{3 \times 10^{-11}}{100 \times 19.89 \times 10^{-26}} \approx 1.51 \times 10^{12} \ s^{-1}$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photo-electrons is given by $K_{max} = hf - \phi$,where $\phi = hf_0$ is the work function.
For the first photo-cathode: $\frac{1}{2}mv_1^2 = hf_1 - hf_0$ (Equation $1$)
For the second photo-cathode: $\frac{1}{2}mv_2^2 = hf_2 - hf_0$ (Equation $2$)
Subtracting Equation $2$ from Equation $1$:
$\frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2 = (hf_1 - hf_0) - (hf_2 - hf_0)$
$\frac{1}{2}m(v_1^2 - v_2^2) = h(f_1 - f_2)$
$v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$

(B) According to Einstein's photoelectric equation: $K_{max} = hv - \phi_0$, where $\phi_0 = hv_0$ is the work function.
Also, $K_{max} = eV_0$, so $eV_0 = hv - hv_0$.
This implies $V_0 = \frac{h}{e}v - \frac{hv_0}{e}$.
Statement $2$ is correct because $K_{max}$ and $V_0$ vary linearly with frequency $v$.
For Statement $1$, if frequency $v$ is doubled to $2v$, the new kinetic energy is $K'_{max} = h(2v) - hv_0 = 2hv - hv_0$.
Since $2hv - hv_0 \neq 2(hv - hv_0)$, $K_{max}$ does not double. Similarly, $V_0$ does not double.
Thus, Statement $1$ is false.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ 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 of the metal.
Stopping potential $V_s$ is related to maximum kinetic energy by $K_{max} = eV_s$.
Therefore,$eV_s = h\nu - \Phi_0$,which implies $V_s = \frac{h}{e}\nu - \frac{\Phi_0}{e}$.
This equation shows that the stopping potential $V_s$ is a linear function of the frequency $\nu$ of the incident light.
As the frequency $\nu$ of the incident light increases,the stopping potential $V_s$ also increases.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{\max})$ of the emitted photoelectrons is given by:
$K_{\max} = E - W_0$
Where:
$E$ is the energy of the incident photon = $3.4 \ eV$
$W_0$ is the work function of the metal = $2 \ eV$
Substituting the values:
$K_{\max} = 3.4 \ eV - 2 \ eV = 1.4 \ eV$
Therefore,the maximum kinetic energy of the photoelectrons is $1.4 \ eV$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = h\nu - \phi$,where $h\nu$ is the energy of the incident photon and $\phi$ is the work function of the metal.
Since some electrons lose energy due to collisions before escaping the surface,the actual kinetic energy $K$ of the emitted photoelectrons ranges from $0$ to $K_{max}$.
Therefore,the kinetic energy of any emitted photoelectron is always less than or equal to the maximum kinetic energy $(K \le K_{max})$.

(B) The threshold wavelength is given as $\lambda_{th} = 200 \ nm$.
The energy of the incident photon is $E = \frac{hc}{\lambda}$,where $\lambda = 100 \ nm$.
Using the formula for maximum kinetic energy $K_{max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_{th}}$.
Using the constant $hc \approx 12400 \ eV \cdot nm$:
$K_{max} = \frac{12400}{100} - \frac{12400}{200} \ eV$.
$K_{max} = 124 - 62 = 62 \ eV$.

(B) Given: Intensity $I = 1 \ W/m^2$, Wavelength $\lambda = 5 \times 10^{-7} \ m$, Area $A = 1 \ cm^2 = 10^{-4} \ m^2$.
Power $P = I \times A = 1 \times 10^{-4} \ W = 10^{-4} \ W$.
Energy of one photon $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{5 \times 10^{-7}} \approx 3.978 \times 10^{-19} \ J$.
Number of photons incident per second $n_p = \frac{P}{E} = \frac{10^{-4}}{3.978 \times 10^{-19}} \approx 2.51 \times 10^{14} \ s^{-1}$.
Since $100$ photons emit $1$ electron, the number of electrons emitted per second $n_e = \frac{n_p}{100} = 2.51 \times 10^{12} \ s^{-1}$.
Photoelectric current $I_p = n_e \times e = 2.51 \times 10^{12} \times 1.6 \times 10^{-19} \ C/s \approx 4.0 \times 10^{-7} \ A = 0.4 \ \mu A$.

(D) The work function $\Phi_0$ is given as $2.3 \ eV$.
The threshold wavelength $\lambda_0$ is related to the work function by the formula $\Phi_0 = \frac{hc}{\lambda_0}$.
Using the approximation $hc \approx 1240 \ eV \cdot nm$,we have:
$\lambda_0 = \frac{1240 \ eV \cdot nm}{2.3 \ eV} \approx 539.13 \ nm$.
This value of approximately $540 \ nm$ falls within the visible spectrum,specifically in the yellow region ($570 \ nm - 590 \ nm$ is typically yellow,but $540 \ nm$ is often categorized as green-yellow or yellow-green; among the given options,yellow is the most appropriate choice).

(B, C) The stopping potential depends only on the frequency of the incident light,not on its intensity. Since the source remains the same (monochromatic),the frequency is constant. Therefore,the stopping potential remains $0.6 \ V$.
The saturation photocurrent $(I_s)$ is directly proportional to the intensity of the incident light. The intensity $(I)$ of a point source follows the inverse square law,$I \propto 1/d^2$.
Given the initial distance $d_1 = 0.2 \ m$ and the final distance $d_2 = 0.6 \ m$,the ratio of the distances is $d_2/d_1 = 0.6/0.2 = 3$.
Thus,the new intensity $I_2 = I_1 / (3^2) = I_1 / 9$.
Since $I_s \propto I$,the new saturation current $I_{s2} = I_{s1} / 9 = 18 \ mA / 9 = 2 \ mA$.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using the approximation $E \approx \frac{12400 \ \text{eV} \cdot \mathring A}{\lambda (\mathring A)}$,we get:
$E = \frac{12400}{4100} \approx 3.02 \ eV$.
Photoelectric emission occurs if the energy of the incident photon is greater than the work function $(\Phi)$ of the metal.
For metal $A$: $\Phi_A = 1.92 \ eV < 3.02 \ eV$ (Emission occurs).
For metal $B$: $\Phi_B = 2.0 \ eV < 3.02 \ eV$ (Emission occurs).
For metal $C$: $\Phi_C = 5.0 \ eV > 3.02 \ eV$ (No emission).
Therefore,photoelectrons will be emitted from metals $A$ and $B$ only.

(B) The work function $\Phi_0$ is given by the formula $\Phi_0 = h \nu_0$,where $h$ is Planck's constant and $\nu_0$ is the threshold frequency.
Given: $\nu_0 = 3 \times 10^{14} \ Hz$ and $h \approx 6.63 \times 10^{-34} \ J \cdot s$.
Calculating the work function:
$\Phi_0 = (6.63 \times 10^{-34} \ J \cdot s) \times (3 \times 10^{14} \ Hz)$
$\Phi_0 = 19.89 \times 10^{-20} \ J$
$\Phi_0 \approx 2.0 \times 10^{-19} \ J$.
Thus,the correct option is $B$.

(B) The condition for photoelectric emission is that the incident light must have a wavelength $\lambda$ less than or equal to the threshold wavelength $\lambda_{Th}$.
Given: $\lambda_{Th} = 5200 \ \mathring A = 520 \ nm$.
$IR$ (Infrared) radiation typically has wavelengths greater than $700 \ nm$,which is much larger than $520 \ nm$.
$UV$ (Ultraviolet) radiation typically has wavelengths in the range of $100 \ nm$ to $400 \ nm$.
Since the $UV$ lamp provides radiation with $\lambda < 520 \ nm$,it satisfies the condition for photoelectric emission.
Therefore,the $50 \ W$ $UV$ lamp will cause photoelectric emission.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by:
$K = hv - E_0$
where $h$ is Planck's constant,$v$ is the frequency of incident radiation,and $E_0$ is the work function.
When the frequency of incident radiation is doubled,the new frequency becomes $v' = 2v$.
The new maximum kinetic energy $K'$ is:
$K' = h(2v) - E_0$
$K' = 2hv - E_0$
We can rewrite this expression by adding and subtracting $hv$:
$K' = hv + hv - E_0$
Since $K = hv - E_0$,we substitute this into the equation:
$K' = hv + K$
Therefore,the new maximum kinetic energy is $K + hv$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by:
$K_{max} = h\nu - \phi_0$,where $h$ is Planck's constant,$\nu$ is the frequency of incident radiation,and $\phi_0$ is the work function of the metal.
Since $\phi_0$ is a constant for a given metal,$K_{max}$ depends only on the frequency $(\nu)$ of the incident radiation.
It does not depend on the intensity of the radiation.

(A) Comparing Einstein's photoelectric equation $\frac{1}{2}mv_{\max}^2 = hf - \phi_0$ with the linear equation $y = mx + c$,
where $y = \frac{1}{2}mv_{\max}^2$,$x = f$,$m = h$,and $c = -\phi_0$.
The slope of the graph is Planck's constant $h$,which is a universal constant and is independent of the nature of the metal or the intensity of the incident radiation.

(A) Given: Threshold wavelength $\lambda_0 = 400 \ nm = 4000 \ \mathring{A}$.
Maximum kinetic energy $K.E._{\max} = 1.5 \ eV$.
Using Einstein's photoelectric equation: $K.E._{\max} = \frac{12400 \ eV \cdot \mathring{A}}{\lambda} - \frac{12400 \ eV \cdot \mathring{A}}{\lambda_0}$.
Substituting the values: $1.5 \ eV = \frac{12400 \ eV \cdot \mathring{A}}{\lambda} - \frac{12400 \ eV \cdot \mathring{A}}{4000 \ \mathring{A}}$.
$1.5 \ eV = \frac{12400 \ eV \cdot \mathring{A}}{\lambda} - 3.1 \ eV$.
$1.5 \ eV + 3.1 \ eV = \frac{12400 \ eV \cdot \mathring{A}}{\lambda}$.
$4.6 \ eV = \frac{12400 \ eV \cdot \mathring{A}}{\lambda}$.
$\lambda = \frac{12400}{4.6} \ \mathring{A} \approx 2695.65 \ \mathring{A} \approx 2700 \ \mathring{A}$.

(A) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + K_{max}$,where $K_{max} = \frac{1}{2}mv^2$.
For $\lambda_1 = 400 \ nm = 400 \times 10^{-9} \ m$ and velocity $v_1 = v$:
$\frac{hc}{400 \times 10^{-9}} = W_0 + \frac{1}{2}mv^2 \quad \dots(i)$
For $\lambda_2 = 250 \ nm = 250 \times 10^{-9} \ m$ and velocity $v_2 = 2v$:
$\frac{hc}{250 \times 10^{-9}} = W_0 + \frac{1}{2}m(2v)^2 = W_0 + 2mv^2 \quad \dots(ii)$
From $(i)$,$\frac{1}{2}mv^2 = \frac{hc}{400 \times 10^{-9}} - W_0$. Substituting this into $(ii)$:
$\frac{hc}{250 \times 10^{-9}} = W_0 + 4 \left( \frac{hc}{400 \times 10^{-9}} - W_0 \right)$
$\frac{hc}{250 \times 10^{-9}} = W_0 + \frac{hc}{100 \times 10^{-9}} - 4W_0$
$3W_0 = hc \left( \frac{1}{100 \times 10^{-9}} - \frac{1}{250 \times 10^{-9}} \right)$
$3W_0 = hc \left( \frac{2.5 - 1}{250 \times 10^{-9}} \right) = hc \left( \frac{1.5}{250 \times 10^{-9}} \right) = hc \left( \frac{15}{2500 \times 10^{-9}} \right) = hc \left( \frac{3}{500 \times 10^{-9}} \right)$
$W_0 = \frac{hc}{500 \times 10^{-9}} = 0.2 \times 10^7 \ hc = 2 \times 10^6 \ hc \ J$.

(A) According to Einstein's photoelectric equation, the maximum kinetic energy 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.
The maximum kinetic energy $(K_{max})$ depends only on the frequency of the incident light and the work function of the metal surface.
It does not depend on the intensity of the incident light.
Therefore, if the intensity is reduced to $1/4$ of its original value, the maximum kinetic energy of the emitted photoelectrons remains constant.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = E - W_0$,where $E$ is the energy of the incident photon and $W_0$ is the work function of the metal.
For the first beam: $E_1 = 1 \ eV$,$W_0 = 0.5 \ eV$.
$K_1 = E_1 - W_0 = 1 \ eV - 0.5 \ eV = 0.5 \ eV$.
For the second beam: $E_2 = 2.5 \ eV$,$W_0 = 0.5 \ eV$.
$K_2 = E_2 - W_0 = 2.5 \ eV - 0.5 \ eV = 2.0 \ eV$.
The ratio of the maximum kinetic energies is $\frac{K_1}{K_2} = \frac{0.5 \ eV}{2.0 \ eV} = \frac{1}{4}$.
Thus,the ratio is $1 : 4$.

(D) According to Einstein's photoelectric equation:
$K_{max} = \frac{1}{2} m u^2 = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function.
For wavelength $\lambda$: $\frac{1}{2} m u^2 = \frac{hc}{\lambda} - \phi$ ---$(1)$
For wavelength $\lambda' = \frac{3\lambda}{4}$: $\frac{1}{2} m u_1^2 = \frac{hc}{\lambda'} - \phi = \frac{4hc}{3\lambda} - \phi$ ---$(2)$
From $(1)$,$\frac{hc}{\lambda} = \frac{1}{2} m u^2 + \phi$.
Substitute this into $(2)$:
$\frac{1}{2} m u_1^2 = \frac{4}{3} (\frac{1}{2} m u^2 + \phi) - \phi$
$\frac{1}{2} m u_1^2 = \frac{4}{3} (\frac{1}{2} m u^2) + \frac{4}{3} \phi - \phi$
$\frac{1}{2} m u_1^2 = \frac{4}{3} (\frac{1}{2} m u^2) + \frac{1}{3} \phi$
Since $\phi > 0$,it follows that $\frac{1}{2} m u_1^2 > \frac{4}{3} (\frac{1}{2} m u^2)$.
$u_1^2 > \frac{4}{3} u^2 \Rightarrow u_1 > \sqrt{\frac{4}{3}} u$.

(C) Einstein's photoelectric equation is given by $K_{max} = h\nu - \phi_0$,where $K_{max}$ is the maximum kinetic energy,$h$ is Planck's constant,$\nu$ is the frequency of incident radiation,and $\phi_0 = h\nu_0$ 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 $c = -\phi_0$ (y-intercept).
Since the y-intercept is negative $(-\phi_0)$,the graph starts from the x-axis at a frequency $\nu = \nu_0$ (threshold frequency) and increases linearly with a positive slope $h$. Thus,the graph has a positive x-intercept at $\nu_0$.

(A) According to Einstein's photoelectric equation,the energy of the incident photon is given by $E = h\nu$.
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.
Given the incident frequency $\nu = 4\nu_0$,the energy of the incident photon is $E = h(4\nu_0) = 4h\nu_0$.
Substituting these values into the equation:
$4h\nu_0 = h\nu_0 + K_{\max}$
Solving for $K_{\max}$:
$K_{\max} = 4h\nu_0 - h\nu_0 = 3h\nu_0$.