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

(B) The work function is given as $W_0 = 3.31 \times 10^{-19} \,J$.
The wavelength of incident radiation is $\lambda = 5000 \text{ Å} = 5000 \times 10^{-10} \,m = 5 \times 10^{-7} \,m$.
According to Einstein's photoelectric equation,the energy of the incident photon $E$ is given by $E = W_0 + KE_{max}$,where $KE_{max}$ is the maximum kinetic energy.
The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{5 \times 10^{-7}} \,J$.
$E = \frac{19.86 \times 10^{-26}}{5 \times 10^{-7}} = 3.972 \times 10^{-19} \,J$.
Now,$KE_{max} = E - W_0 = 3.972 \times 10^{-19} \,J - 3.31 \times 10^{-19} \,J = 0.662 \times 10^{-19} \,J$.
To convert this energy into electron-volts $(eV)$,divide by the charge of an electron $e = 1.6 \times 10^{-19} \,C$:
$KE_{max} = \frac{0.662 \times 10^{-19} \,J}{1.6 \times 10^{-19} \,C} = 0.41375 \,eV \approx 0.41 \,eV$.

(C) According to Einstein's photoelectric equation: $e V_0 = h \nu - h \nu_0$,where $V_0$ is the stopping potential,$\nu$ is the frequency of incident light,and $\nu_0$ is the threshold frequency.
Rearranging for the threshold frequency $\nu_0$: $\nu_0 = \nu - \frac{e V_0}{h}$.
Since $\nu = \frac{c}{\lambda}$,we have $\nu_0 = \frac{c}{\lambda} - \frac{e V_0}{h}$.
Substituting the given values: $c = 3 \times 10^8 \,m/s$,$\lambda = 2 \times 10^{-7} \,m$,$e = 1.6 \times 10^{-19} \,C$,$V_0 = 2.5 \,V$,and $h = 6.6 \times 10^{-34} \,J-s$.
$\nu_0 = \frac{3 \times 10^8}{2 \times 10^{-7}} - \frac{1.6 \times 10^{-19} \times 2.5}{6.6 \times 10^{-34}}$
$\nu_0 = 1.5 \times 10^{15} - 0.606 \times 10^{15} \approx 0.894 \times 10^{15} \,Hz$.
Rounding to the nearest significant value,$\nu_0 \approx 9.0 \times 10^{14} \,Hz$.

(C) The stopping potential $V_0$ is related to the incident energy and work function by the Einstein's photoelectric equation: $e V_0 = K_{\text{max}} = \frac{hc}{\lambda} - \phi$.
Given:
Work function $\phi = 5 \text{ eV} = 5 \times 1.6 \times 10^{-19} \text{ J} = 8 \times 10^{-19} \text{ J}$.
Wavelength $\lambda = 2000 \text{ Å} = 2 \times 10^{-7} \text{ m}$.
Planck's constant $h = 6.67 \times 10^{-34} \text{ J-s}$.
Speed of light $c = 3 \times 10^8 \text{ m/s}$.
Calculating the energy of incident photon:
$E = \frac{hc}{\lambda} = \frac{6.67 \times 10^{-34} \times 3 \times 10^8}{2 \times 10^{-7}} = 10.005 \times 10^{-19} \text{ J}$.
Now,substituting into the equation:
$e V_0 = 10.005 \times 10^{-19} \text{ J} - 8 \times 10^{-19} \text{ J} = 2.005 \times 10^{-19} \text{ J}$.
$V_0 = \frac{2.005 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ C}} \approx 1.25 \text{ V}$.

(C) Einstein's photoelectric equation is given by:
$h \nu = h \nu_0 + K_{max}$
Since the stopping potential $V_0 = 3 \ V$,the maximum kinetic energy $K_{max} = e V_0 = 1.6 \times 10^{-19} \times 3 \ J$.
The threshold frequency $\nu_0 = 6 \times 10^{14} \ s^{-1}$.
Using the relation $h \nu = h \nu_0 + e V_0$,we get:
$\nu = \nu_0 + \frac{e V_0}{h}$
$\nu = 6 \times 10^{14} + \frac{1.6 \times 10^{-19} \times 3}{6.4 \times 10^{-34}}$
$\nu = 6 \times 10^{14} + \frac{4.8 \times 10^{-19}}{6.4 \times 10^{-34}}$
$\nu = 6 \times 10^{14} + 0.75 \times 10^{15}$
$\nu = 6 \times 10^{14} + 7.5 \times 10^{14} = 13.5 \times 10^{14} \ s^{-1}$.

(B) The Einstein's photoelectric equation is given by $eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $V_0$ is the stopping potential and $\lambda_0$ is the threshold wavelength.
For the first case: $4.8 = \frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \quad \dots (i)$
For the second case: $1.6 = \frac{hc}{e} \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) \quad \dots (ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{4.8}{1.6} = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$
$3 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$
$3 \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{3}{2\lambda} - \frac{3}{\lambda_0} = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{3}{2\lambda} - \frac{1}{\lambda} = \frac{3}{\lambda_0} - \frac{1}{\lambda_0}$
$\frac{1}{2\lambda} = \frac{2}{\lambda_0}$
$\lambda_0 = 4\lambda$

(D) Let the work function of the metal be $W$.
The energy of the first photon is $E_1 = 2W$.
The energy of the second photon is $E_2 = 3W$.
According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max}$ is given by $(KE)_{\max} = E - W$.
For the first photon: $(KE_1)_{\max} = 2W - W = W$.
For the second photon: $(KE_2)_{\max} = 3W - W = 2W$.
The ratio of maximum kinetic energies is $\frac{(KE_1)_{\max}}{(KE_2)_{\max}} = \frac{W}{2W} = \frac{1}{2}$.
Since $(KE)_{\max} = \frac{1}{2}mv^2$,we have $\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{1}{2}$.
This simplifies to $\frac{v_1^2}{v_2^2} = \frac{1}{2}$,which gives $\frac{v_1}{v_2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Thus,the ratio of maximum velocities is $1 : \sqrt{2}$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max}$ is given by:
$(KE)_{\max} = h v - h v_0$,where $v$ is the frequency of incident radiation and $v_0$ is the threshold frequency.
For the first case with frequency $v_1$: $(KE_1)_{\max} = h(v_1 - v_0)$.
For the second case with frequency $v_2$: $(KE_2)_{\max} = h(v_2 - v_0)$.
Given the ratio of maximum kinetic energies is $1:k$,we have:
$\frac{(KE_1)_{\max}}{(KE_2)_{\max}} = \frac{1}{k} = \frac{h(v_1 - v_0)}{h(v_2 - v_0)}$.
Cross-multiplying gives:
$v_2 - v_0 = k(v_1 - v_0)$.
$v_2 - v_0 = k v_1 - k v_0$.
Rearranging to solve for $v_0$:
$k v_0 - v_0 = k v_1 - v_2$.
$v_0(k - 1) = k v_1 - v_2$.
$v_0 = \frac{k v_1 - v_2}{k - 1}$.

(C) Albert Einstein is widely known for his work on the special theory of relativity and general relativity. However,he was awarded the Nobel Prize in Physics in $1921$ specifically for his discovery of the law of the photoelectric effect.
Therefore,the correct option is $C$.

(C) Given: Energy of incident photons $E = 10.5 \ eV$.
Maximum velocity of photoelectrons $v = 1.6 \times 10^6 \ m \ s^{-1}$.
Mass of electron $m = 9 \times 10^{-31} \ kg$.
The maximum kinetic energy $K.E._{\max}$ is given by $\frac{1}{2}mv^2$.
$K.E._{\max} = \frac{1}{2} \times (9 \times 10^{-31}) \times (1.6 \times 10^6)^2 = 0.5 \times 9 \times 10^{-31} \times 2.56 \times 10^{12} = 11.52 \times 10^{-19} \ J$.
Converting this to electron-volts: $K.E._{\max} = \frac{11.52 \times 10^{-19}}{1.6 \times 10^{-19}} \ eV = 7.2 \ eV$.
According to Einstein's photoelectric equation,$E = \phi + K.E._{\max}$,where $\phi$ is the work function.
$\phi = E - K.E._{\max} = 10.5 \ eV - 7.2 \ eV = 3.3 \ eV$.

(C) The energy of the photon emitted during the transition from $n=2$ to $n=1$ is given by:
$E = E_2 - E_1 = -\frac{13.6}{2^2} - (-\frac{13.6}{1^2}) = -3.4 + 13.6 = 10.2 \ eV$.
According to Einstein's photoelectric equation:
$E = \phi + K_{max}$,where $\phi$ is the work function and $K_{max} = eV_s$ is the maximum kinetic energy.
$10.2 \ eV = 4.2 \ eV + eV_s$.
$eV_s = 10.2 \ eV - 4.2 \ eV = 6.0 \ eV$.
Therefore,the stopping potential $V_s = 6 \ V$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ of the emitted electrons is given by:
$K = \frac{hc}{\lambda} - \phi$
When an electron of charge $q$ and mass $m$ moves perpendicular to a magnetic field $B$,it follows a circular path of radius $R$ given by:
$R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$
Squaring both sides,we get:
$R^2 = \frac{2mK}{q^2 B^2}$
Rearranging for $B^2$:
$B^2 = \frac{2mK}{q^2 R^2}$
Substituting the expression for $K$:
$B^2 = \frac{2m}{q^2 R^2} \left( \frac{hc}{\lambda} - \phi \right)$
$B^2 = \left( \frac{2mhc}{q^2 R^2} \right) \frac{1}{\lambda} - \left( \frac{2m\phi}{q^2 R^2} \right)$
This is an equation of a straight line of the form $y = mx + c$,where $y = B^2$,$x = \frac{1}{\lambda}$,slope $m = \frac{2mhc}{q^2 R^2}$ (positive),and intercept $c = -\frac{2m\phi}{q^2 R^2}$ (negative).
Thus,the graph is a straight line with a positive slope and a negative y-intercept. This corresponds to Graph $C$.

(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $hc \approx 12400 \ \text{eV} \cdot \mathring{A}$,we get $E = \frac{12400}{4770} \ \text{eV} \approx 2.6 \ \text{eV}$.
For photoelectric emission to occur,the energy of the incident photon must be greater than or equal to the work function $(\phi)$ of the metal $(E \ge \phi)$.
Comparing $E = 2.6 \ \text{eV}$ with the work functions:
For metal $A$: $2.6 \ \text{eV} < 4.2 \ \text{eV}$ (No emission)
For metal $B$: $2.6 \ \text{eV} < 3.7 \ \text{eV}$ (No emission)
For metal $C$: $2.6 \ \text{eV} < 3.2 \ \text{eV}$ (No emission)
For metal $D$: $2.6 \ \text{eV} > 2.3 \ \text{eV}$ (Emission occurs)
Therefore,electrons will be emitted only from metal $D$.

(A, B) The given electric field is $E = a \cos \omega_{0} t + \frac{a}{2} [\sin(\omega + \omega_{0})t + \sin(\omega - \omega_{0})t]$.
This indicates the presence of three frequencies: $f_{0} = \frac{\omega_{0}}{2\pi}$,$f_{1} = \frac{\omega + \omega_{0}}{2\pi}$,and $f_{2} = \frac{|\omega - \omega_{0}|}{2\pi}$.
The maximum energy corresponds to the highest frequency $f_{1} = \frac{6 \times 10^{15}}{2\pi} \text{ Hz}$.
The energy of this photon is $E_{max} = h f_{1} = \frac{6.62 \times 10^{-34} \times 6 \times 10^{15}}{2 \times 3.14 \times 1.6 \times 10^{-19}} \text{ eV} \approx 3.95 \text{ eV}$.
Given stopping potential $V_{s} = 2 \text{ V}$,so $KE_{max} = 2 \text{ eV}$.
Using $KE_{max} = E_{max} - \phi$,we get $2 \text{ eV} = 3.95 \text{ eV} - \phi$,so $\phi = 1.95 \text{ eV}$.
For frequency $\omega = 10^{15} \text{ s}^{-1}$,energy $E = h(\frac{\omega}{2\pi}) \approx 0.66 \text{ eV}$.
Since $E < \phi$,the photoelectric effect is not possible for frequency $\omega$.
Einstein's photoelectric equation $eV_{s} = hf - \phi$ represents a straight line graph between $V_{s}$ and $f$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $T$ of the emitted photoelectrons is given by $T = hv - \phi$,where $h$ is Planck's constant,$v$ is the frequency of incident light,and $\phi$ is the work function of the metal.
From this equation,it is clear that $T$ increases linearly with the frequency $v$ of the incident light.
Additionally,the number of photoelectrons emitted per second is directly proportional to the intensity $J$ of the incident light,provided the frequency $v$ is greater than the threshold frequency $v_{0}$.
Since both statements $(B)$ and $(D)$ are physically correct,in the context of standard multiple-choice questions where only one answer is expected,$(D)$ is often cited as a fundamental observation regarding intensity,while $(B)$ describes the energy dependence. Given the options,$(D)$ is a standard result of the photoelectric effect.

(C) From Einstein's photoelectric equation,$h v = W + e V$,where $W$ is the work function,$v$ is the frequency of incident light,and $V$ is the stopping potential.
For frequency $v_{1}$,we have: $h v_{1} = W + e V_{1}$ --- $(i)$
For frequency $v_{2}$,we have: $h v_{2} = W + e V_{2}$ --- (ii)
Subtracting equation $(i)$ from equation (ii):
$h(v_{2} - v_{1}) = e(V_{2} - V_{1})$
$e = \frac{h(v_{2} - v_{1})}{V_{2} - V_{1}}$
However,we need to express $e$ in terms of $W$. From $(i)$,$h = \frac{W + e V_{1}}{v_{1}}$. Substituting this into (ii):
$(\frac{W + e V_{1}}{v_{1}}) v_{2} = W + e V_{2}$
$W v_{2} + e V_{1} v_{2} = W v_{1} + e V_{2} v_{1}$
$e(V_{1} v_{2} - V_{2} v_{1}) = W v_{1} - W v_{2}$
$e(V_{1} v_{2} - V_{2} v_{1}) = -W(v_{2} - v_{1})$
$e = \frac{W(v_{2} - v_{1})}{V_{2} v_{1} - V_{1} v_{2}}$
This matches option $C$.

(D) The work function of Cesium is $\phi = 2.27 eV$.
The wavelength of the incident light is $\lambda = 600 nm = 600 \times 10^{-9} m$.
The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Substituting the values ($h = 6.63 \times 10^{-34} Js$,$c = 3 \times 10^8 m/s$):
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}} J = 3.315 \times 10^{-19} J$.
Converting this energy into electron-volts $(eV)$:
$E = \frac{3.315 \times 10^{-19}}{1.6 \times 10^{-19}} eV \approx 2.07 eV$.
Since the energy of the incident photon $(2.07 eV)$ is less than the work function of Cesium $(2.27 eV)$,the photoelectric effect will not occur.
Therefore,no electrons are emitted,and no cut-off voltage is required. Thus,the correct answer is $D$.

(B) The photoelectric effect occurs when the energy of incident photons $E = \frac{hc}{\lambda}$ is greater than or equal to the work function $\phi$ of the metal.
Given the range of work function $2 eV \leq \phi \leq 5 eV$,the threshold wavelength $\lambda$ must satisfy $\lambda \leq \frac{hc}{\phi}$.
For $\phi = 2 eV$,$\lambda_{\max} = \frac{4 \times 10^{-15} eVs \times 3 \times 10^{8} m/s}{2 eV} = 6 \times 10^{-7} m = 600 nm$.
For $\phi = 5 eV$,$\lambda_{\min} = \frac{4 \times 10^{-15} eVs \times 3 \times 10^{8} m/s}{5 eV} = 2.4 \times 10^{-7} m = 240 nm$.
Thus,the range of wavelengths that can cause the photoelectric effect is $240 nm \leq \lambda \leq 600 nm$.
Comparing the given options,$650 nm$ is greater than $600 nm$,so it cannot cause the photoelectric effect.

(A, B, D) According to the experimental observations of the photoelectric effect:
$1$. There is no significant time delay between the absorption of suitable radiation and the emission of electrons,even at very low intensities.
$2$. Einstein's photoelectric equation is $K_{max} = h\nu - \phi_0$,where $\phi_0$ is the work function. Electrons are emitted only if $\nu > \nu_0$ (threshold frequency). Thus,no electrons are emitted below the threshold frequency.
$3$. The maximum kinetic energy $K_{max}$ is a linear function of frequency $\nu$,but it is not directly proportional to it (due to the work function term $\phi_0$).
$4$. The maximum kinetic energy $K_{max}$ depends only on the frequency of incident radiation and the work function of the metal; it is independent of the intensity of the incident radiation.
Therefore,statements $(a)$,$(b)$,and $(d)$ are correct.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of photoelectrons is given by $K_{max} = h v - \phi_{0}$,where $\phi_{0}$ is the work function of the metal.
Since $K_{max} = e V_{s}$,where $V_{s}$ is the stopping potential,we have $e V_{s} = h v - \phi_{0}$.
For the first case: $e V_{1} = h v_{1} - \phi_{0}$ ---$(i)$
For the second case: $e V_{2} = h v_{2} - \phi_{0}$ ---(ii)
Subtracting equation $(i)$ from equation (ii):
$e V_{2} - e V_{1} = (h v_{2} - \phi_{0}) - (h v_{1} - \phi_{0})$
$e(V_{2} - V_{1}) = h(v_{2} - v_{1})$
$v_{2} - v_{1} = \frac{e}{h}(V_{2} - V_{1})$
$v_{2} = v_{1} + \frac{e}{h}(V_{2} - V_{1})$

(B) According to Einstein's photoelectric equation: $h\nu = h\nu_{0} + eV_{0}$,where $\nu_{0}$ is the threshold frequency.
For the first case:
$h\nu = h\nu_{0} + eV_{0}$ --- $(i)$
For the second case:
$h(\frac{\nu}{2}) = h\nu_{0} + e(\frac{V_{0}}{4})$ --- (ii)
From equation $(i)$,we have $eV_{0} = h\nu - h\nu_{0}$.
Substitute this into equation (ii):
$\frac{h\nu}{2} = h\nu_{0} + \frac{1}{4}(h\nu - h\nu_{0})$
Multiply the entire equation by $4$ to clear the denominator:
$2h\nu = 4h\nu_{0} + h\nu - h\nu_{0}$
$2h\nu = 3h\nu_{0} + h\nu$
$h\nu = 3h\nu_{0}$
$\nu_{0} = \frac{\nu}{3}$

(A) According to Einstein's photoelectric equation,the energy of the incident photon is equal to the work function when the kinetic energy of the emitted photoelectrons is zero.
$ E = h\nu = \phi $
Where $ \phi = 110 \times 10^{-20} \ J $ is the work function and $ h = 6.63 \times 10^{-34} \ J \cdot s $ is Planck's constant.
The frequency $ \nu $ is given by $ \nu = \frac{\phi}{h} $.
The angular frequency $ \omega $ is related to frequency by $ \omega = 2\pi\nu $.
Substituting the values:
$ \omega = 2\pi \left( \frac{\phi}{h} \right) = \frac{2 \times 3.14 \times 110 \times 10^{-20}}{6.63 \times 10^{-34}} $
$ \omega \approx 1.04 \times 10^{16} \ rad/s $.

(A) The given electric field equation is $E = 60 \sin(3 \times 10^{15} t) + \sin(12 \times 10^{15} t)$.
This represents two light waves with angular frequencies $\omega_1 = 3 \times 10^{15} \text{ rad/s}$ and $\omega_2 = 12 \times 10^{15} \text{ rad/s}$.
The energy of a photon is given by $E_{ph} = \hbar \omega = \frac{h \omega}{2\pi}$.
For the higher frequency component $\omega_2 = 12 \times 10^{15} \text{ rad/s}$:
$E_{ph} = \frac{6.63 \times 10^{-34} \times 12 \times 10^{15}}{2 \times 3.14} \text{ J}$.
$E_{ph} \approx 1.265 \times 10^{-18} \text{ J}$.
Converting this to electron-volts $(\text{eV})$: $E_{ph} = \frac{1.265 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 7.9 \text{ eV}$.
According to Einstein's photoelectric equation,$K_{max} = E_{ph} - \phi_0$.
Given work function $\phi_0 = 2.8 \text{ eV}$.
$K_{max} = 7.9 \text{ eV} - 2.8 \text{ eV} = 5.1 \text{ eV}$.

(D) Einstein's photoelectric equation is given by $eV_0 = h\nu - \Phi$,where $\Phi$ is the work function.
Rearranging the equation to the form $y = mx + c$,we get $V_0 = (\frac{h}{e})\nu - \frac{\Phi}{e}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = V_0$ and $x = \nu$,the slope $m$ is equal to $\frac{h}{e}$.
Therefore,the correct option is $D$.

(B) The maximum kinetic energy $(K_{max})$ of photoelectrons is related to the stopping potential $(V_0)$ by the equation $K_{max} = eV_0$.
Given that the cut-off voltage (stopping potential) $V_0 = 1.5 \text{ V}$.
Substituting the value into the equation,we get $K_{max} = e \times 1.5 \text{ V} = 1.5 \text{ eV}$.
Therefore,the maximum kinetic energy is $1.5 \text{ eV}$.

(A) According to Einstein's photoelectric equation,$K_{max} = h\nu - \phi = \frac{hc}{\lambda} - \phi$.
For the same wavelength $\lambda$,the energy of the incident photon $\frac{hc}{\lambda}$ is constant.
The maximum kinetic energy $K_{max}$ is higher for the metal with the smaller work function $\phi$.
The work function is given by $\phi = h\nu_0$,where $\nu_0$ is the threshold frequency (the $x$-intercept of the graph).
From the graph,$X_1$ has the lowest threshold frequency $(\nu_0 = 1.0 \times 10^{14} \text{ Hz})$,which implies it has the smallest work function.
Therefore,for a fixed incident wavelength,metal $X_1$ will emit photoelectrons with the greatest maximum kinetic energy.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Substituting the given values: $E = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9}} = \frac{19.86 \times 10^{-26}}{331 \times 10^{-9}} = 0.06 \times 10^{-17} \text{ J} = 6 \times 10^{-19} \text{ J}$.
According to Einstein's photoelectric equation,$E = \phi + K_{max}$,where $\phi$ is the work function and $K_{max}$ is the maximum kinetic energy.
The maximum kinetic energy is $K_{max} = eV_s = 1.6 \times 10^{-19} \times 0.2 = 0.32 \times 10^{-19} \text{ J}$.
Therefore,the work function $\phi = E - K_{max} = 6 \times 10^{-19} - 0.32 \times 10^{-19} = 5.68 \times 10^{-19} \text{ J}$.
Comparing this with $\alpha \times 10^{-19} \text{ J}$,we get $\alpha = 5.68$.

(D) 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 wavelengths $\lambda_1$ and $\lambda_2$,we have:
$K_1 = \frac{hc}{\lambda_1} - \phi$ --- $(1)$
$K_2 = \frac{hc}{\lambda_2} - \phi$ --- $(2)$
Given $\lambda_1 = 2\lambda_2$,substitute this into equation $(1)$:
$K_1 = \frac{hc}{2\lambda_2} - \phi$
Multiply by $2$:
$2K_1 = \frac{hc}{\lambda_2} - 2\phi$ --- $(3)$
Now,subtract equation $(3)$ from equation $(2)$:
$K_2 - 2K_1 = (\frac{hc}{\lambda_2} - \phi) - (\frac{hc}{\lambda_2} - 2\phi)$
$K_2 - 2K_1 = \phi$
Therefore,the work function $\phi = K_2 - 2K_1$.

(B) According to Einstein's photoelectric equation: $eV_s = \frac{hc}{\lambda} - \Phi$,where $\Phi$ is the work function.
For the first case: $e(3V_0) = \frac{hc}{\lambda} - \Phi$ --- $(1)$
For the second case: $e(V_0) = \frac{hc}{2\lambda} - \Phi$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$3eV_0 - eV_0 = (\frac{hc}{\lambda} - \Phi) - (\frac{hc}{2\lambda} - \Phi)$
$2eV_0 = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda}$
$eV_0 = \frac{hc}{4\lambda}$
Substitute $eV_0$ into equation $(2)$:
$\frac{hc}{4\lambda} = \frac{hc}{2\lambda} - \Phi$
$\Phi = \frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{hc}{4\lambda}$
The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \frac{hc}{\Phi}$.
Substituting $\Phi = \frac{hc}{4\lambda}$:
$\lambda_0 = \frac{hc}{hc / 4\lambda} = 4\lambda$.
Comparing this with $\alpha\lambda$,we get $\alpha = 4$.

(A) The threshold wavelength $\lambda_0$ is given by the formula $\lambda_0 = \frac{hc}{\phi}$.
Given work function $\phi = 6.6 \text{eV} = 6.6 \times 1.6 \times 10^{-19} \text{J}$.
Substituting the values: $\lambda_0 = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.6 \times 1.6 \times 10^{-19}} \text{m}$.
$\lambda_0 = \frac{3 \times 10^{-26}}{1.6 \times 10^{-19}} \text{m} = 1.875 \times 10^{-7} \text{m} = 187.5 \text{nm}$.
The photoelectric effect occurs only when the incident wavelength $\lambda \le \lambda_0$.
If $\lambda > \lambda_0$, the energy of the incident photon is less than the work function, and no photoelectric emission occurs.
Comparing the options: $200 \text{nm} > 187.5 \text{nm}$, $100 \text{nm} < 187.5 \text{nm}$, $50 \text{nm} < 187.5 \text{nm}$, and $150 \text{nm} < 187.5 \text{nm}$.
Therefore, $200 \text{nm}$ radiation will not cause the photoelectric effect.