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

(C) The energy of an incident photon is given by the formula $E = \frac{hc}{\lambda}$.
Using $hc \approx 12400 \, eV \cdot Å$, we have $E = \frac{12400}{4000} \, eV = 3.1 \, eV$.
Photoelectrons are emitted if the energy of the incident light is greater than or equal to the work function $(\Phi)$ of the material.
Comparing $E = 3.1 \, eV$ with the given work functions:
For $A$: $1.2 \, eV < 3.1 \, eV$ (Emits)
For $B$: $2.4 \, eV < 3.1 \, eV$ (Emits)
For $C$: $3.6 \, eV > 3.1 \, eV$ (Does not emit)
For $D$: $4.8 \, eV > 3.1 \, eV$ (Does not emit)
For $E$: $6.0 \, eV > 3.1 \, eV$ (Does not emit)
Therefore, only elements $A$ and $B$ will emit photoelectrons.

(A) Given:
Energy of incident photon,$E = 8\,eV$.
Threshold frequency,$\nu_0 = 1.6 \times 10^{15}\,Hz$.
Planck's constant,$h = 6.6 \times 10^{-34}\,J\cdot s$.
Conversion factor,$1\,eV = 1.6 \times 10^{-19}\,J$.
The work function $\Phi$ is given by $\Phi = h\nu_0$.
$\Phi = (6.6 \times 10^{-34}\,J\cdot s) \times (1.6 \times 10^{15}\,Hz) = 10.56 \times 10^{-19}\,J$.
Converting the work function into $eV$:
$\Phi = \frac{10.56 \times 10^{-19}\,J}{1.6 \times 10^{-19}\,J/eV} = 6.6\,eV$.
According to Einstein's photoelectric equation,the maximum kinetic energy $KE_{\max}$ is:
$KE_{\max} = E - \Phi$.
$KE_{\max} = 8\,eV - 6.6\,eV = 1.4\,eV$.

(D) According to Einstein's photoelectric equation,$K_{max} = \frac{hc}{\lambda} - \phi_0$,where $K_{max} = \frac{1}{2}mv^2$.
Let $v_1$ and $v_2$ be the maximum velocities for $\lambda_1$ and $\lambda_2$ respectively. Given $v_1 = 2v_2$,so $K_1 = 4K_2$.
$E_1 = \frac{hc}{\lambda_1} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{350 \times 10^{-9}} \approx 5.68 \times 10^{-19} \, J$.
$E_2 = \frac{hc}{\lambda_2} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{450 \times 10^{-9}} \approx 4.42 \times 10^{-19} \, J$.
From the equations: $E_1 - \phi_0 = 4(E_2 - \phi_0)$.
$E_1 - \phi_0 = 4E_2 - 4\phi_0$.
$3\phi_0 = 4E_2 - E_1$.
$3\phi_0 = 4(4.42 \times 10^{-19}) - 5.68 \times 10^{-19} = 17.68 \times 10^{-19} - 5.68 \times 10^{-19} = 12.0 \times 10^{-19} \, J$.
$\phi_0 = 4.0 \times 10^{-19} \, J$.

(A) According to Einstein's photoelectric equation: $K_{\text{max}} = E - \phi_0$,where $E = \frac{hc}{\lambda}$.
Substituting the values: $0.3 \text{ eV} = \frac{hc}{\lambda} - 2.3 \text{ eV}$.
$\frac{hc}{\lambda} = 0.3 \text{ eV} + 2.3 \text{ eV} = 2.6 \text{ eV}$.
Using $hc \approx 12400 \text{ eV} \cdot \mathring A$,we get $\lambda = \frac{12400}{2.6} \mathring A$.
$\lambda \approx 4769.23 \mathring A$.
Rounding to the nearest given option,the wavelength is $4800 \mathring A$.

(A) Einstein's photoelectric equation is given by $\frac{hc}{\lambda} = eV_s + W$,where $V_s$ is the stopping potential and $W$ is the work function.
For the first case: $\frac{hc}{\lambda} = e(5V_0) + W$ --- $(1)$
For the second case: $\frac{hc}{2\lambda} = eV_0 + W$ --- $(2)$
Multiply equation $(2)$ by $5$: $\frac{5hc}{2\lambda} = 5eV_0 + 5W$ --- $(3)$
Subtract equation $(1)$ from equation $(3)$:
$\frac{5hc}{2\lambda} - \frac{hc}{\lambda} = (5eV_0 + 5W) - (5eV_0 + W)$
$\frac{3hc}{2\lambda} = 4W$
$W = \frac{3hc}{8\lambda}$
Since the work function $W = \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength:
$\frac{hc}{\lambda_0} = \frac{3hc}{8\lambda}$
$\lambda_0 = \frac{8}{3}\lambda$

(C) The kinetic energy $(KE)$ of a photoelectron is given by Einstein's photoelectric equation: $KE = h\nu - W = \frac{hc}{\lambda} - W$.
Here,$h$ is Planck's constant,$\nu$ is the frequency,$\lambda$ is the wavelength of incident light,and $W$ is the work function of the metal.
Photoelectrons are emitted from different depths within the metal. As they travel to the surface,they undergo multiple collisions with other electrons and atoms,losing some energy. Therefore,the kinetic energy of emitted photoelectrons varies from $0$ to a maximum value $(KE_{max} = h\nu - W)$.
Thus,the kinetic energy of any emitted photoelectron is always equal to or less than the maximum kinetic energy.

(C) According to Einstein's photoelectric equation, the maximum kinetic energy $(K_{max})$ is given by: $K_{max} = hv - \Phi$, where $h$ is Planck's constant, $v$ is the frequency of incident light, and $\Phi$ is the work function of the metal.
For the first case: $K_0 = hv_1 - \Phi$ --- (Equation $1$)
For the second case: $2K_0 = hv_2 - \Phi$ --- (Equation $2$)
From Equation $1$, we have $\Phi = hv_1 - K_0$. Substituting this into Equation $2$:
$2K_0 = hv_2 - (hv_1 - K_0)$
$2K_0 = hv_2 - hv_1 + K_0$
$K_0 = hv_2 - hv_1$
$hv_2 = hv_1 + K_0$
Since $K_0 = hv_1 - \Phi$, we substitute $K_0$ back:
$hv_2 = hv_1 + (hv_1 - \Phi)$
$hv_2 = 2hv_1 - \Phi$
$v_2 = 2v_1 - \frac{\Phi}{h}$
Since $\Phi > 0$, it follows that $v_2 < 2v_1$.

(C) The maximum kinetic energy for the photoelectrons is given by Einstein's photoelectric equation:
$E_{\max} = h\nu - \phi$
where $\nu$ is the frequency of incident light and $\phi$ is the work function of the metal.
Since $E_{\max} = eV_0$,we have:
$eV_0 = h\nu - \phi$ ... $(1)$
where $V_0$ is the stopping potential.
When the frequency is doubled $(
u' = 2\nu)$,the new stopping potential $V_0'$ is:
$eV_0' = h(2\nu) - \phi = 2h\nu - \phi$
From equation $(1)$,we know $h\nu = eV_0 + \phi$. Substituting this into the expression for $eV_0'$:
$eV_0' = 2(eV_0 + \phi) - \phi$
$eV_0' = 2eV_0 + 2\phi - \phi$
$eV_0' = 2eV_0 + \phi$
Dividing by $e$:
$V_0' = 2V_0 + \frac{\phi}{e}$
Since $\phi > 0$,it is clear that $V_0' > 2V_0$. Therefore,the stopping potential becomes more than double.

(D) According to Einstein's photoelectric equation: $eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0 = 5 \times 10^{-10} \, m$ is the threshold wavelength.
For the first case,$\lambda_1 = 2 \times 10^{-10} \, m$:
$eV_0 = hc \left( \frac{1}{2 \times 10^{-10}} - \frac{1}{5 \times 10^{-10}} \right) = hc \left( \frac{5-2}{10 \times 10^{-10}} \right) = \frac{3hc}{10 \times 10^{-10}}$.
For the second case,the wavelength is doubled: $\lambda_2 = 2 \times \lambda_1 = 4 \times 10^{-10} \, m$.
Let the new stopping potential be $V_0'$.
$eV_0' = hc \left( \frac{1}{4 \times 10^{-10}} - \frac{1}{5 \times 10^{-10}} \right) = hc \left( \frac{5-4}{20 \times 10^{-10}} \right) = \frac{hc}{20 \times 10^{-10}}$.
Comparing $eV_0'$ and $eV_0$:
$\frac{eV_0'}{eV_0} = \frac{hc / (20 \times 10^{-10})}{3hc / (10 \times 10^{-10})} = \frac{1}{20} \times \frac{10}{3} = \frac{1}{6}$.
Therefore,$V_0' = \frac{V_0}{6}$,which is clearly less than $0.5 V_0$.

(D) When a photon of energy $hv$ is absorbed by an electron inside a metal,the electron may lose some of its energy due to collisions with other atoms or electrons before reaching the surface.
If the electron is at the surface and absorbs the photon,it will be emitted with the maximum kinetic energy $K_{max} = hv - \phi$.
However,if the electron is located below the surface,it will lose some energy during its journey to the surface.
Therefore,the electron may either not come out at all (if it loses too much energy) or it may come out with a kinetic energy less than $hv - \phi$.
Thus,the correct statement is that it may come out with kinetic energy less than $hv - \phi$.

(D) According to Einstein's photoelectric equation: $K_{max} = e V_0 = \frac{hc}{\lambda} - \phi$.
Given: Work function $\phi = 2.14 \ eV$,Stopping potential $V_0 = 0.60 \ V$.
Substituting the values: $0.60 \ eV = \frac{1240 \ eV \cdot nm}{\lambda} - 2.14 \ eV$.
$\frac{1240}{\lambda} = 0.60 + 2.14 = 2.74 \ eV$.
$\lambda = \frac{1240}{2.74} \approx 452.55 \ nm$.
Rounding to the nearest integer,the wavelength is approximately $453 \ nm$. Since $453 \ nm$ is not among the options,the correct choice is $D$.

(C) The kinetic energy of the fastest photoelectrons is given by the equation: $\frac{1}{2} mv_{\max}^{2} = eV_s$.
Here,$V_s$ is the stopping potential,$m$ is the mass of the electron,$e$ is the charge of the electron,and $v_{\max}$ is the maximum velocity.
Rearranging the formula for the stopping potential $V_s$:
$V_s = \frac{mv_{\max}^{2}}{2e} = \frac{v_{\max}^{2}}{2(e/m)}$.
Given that the specific charge $(e/m) = 1.8 \times 10^{11}\, C/kg$ and $v_{\max} = 1.2 \times 10^6\, m/s$:
$V_s = \frac{(1.2 \times 10^6)^2}{2 \times 1.8 \times 10^{11}}$.
$V_s = \frac{1.44 \times 10^{12}}{3.6 \times 10^{11}}$.
$V_s = \frac{14.4}{3.6} = 4\, V$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function.
Thus,$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} = hc \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right)$.
So,$v = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right)}$.
When the wavelength is changed to $\lambda' = \frac{3\lambda}{4}$,the new speed $v'$ is:
$v' = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - 3\lambda/4}{(3\lambda/4) \lambda_0} \right)}$.
Taking the ratio:
$\frac{v'}{v} = \sqrt{\frac{\lambda_0 - 3\lambda/4}{3\lambda/4 \lambda_0} \cdot \frac{\lambda \lambda_0}{\lambda_0 - \lambda}} = \sqrt{\frac{4}{3}} \sqrt{\frac{\lambda_0 - 3\lambda/4}{\lambda_0 - \lambda}}$.
Since $\lambda_0 > \lambda$,it follows that $\lambda_0 - 3\lambda/4 > \lambda_0 - \lambda$.
Therefore,$\sqrt{\frac{\lambda_0 - 3\lambda/4}{\lambda_0 - \lambda}} > 1$.
This implies $v' > \sqrt{\frac{4}{3}} v$.

(A) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = h\nu - \phi$,where $\phi$ is the work function.
For frequency $\nu$,the stopping potential is $V_s = \frac{h\nu - \phi}{e}$.
For frequency $\frac{3\nu}{2}$,the stopping potential is $2V_s = \frac{h(\frac{3\nu}{2}) - \phi}{e}$.
Substituting the first equation into the second: $2(\frac{h\nu - \phi}{e}) = \frac{1.5h\nu - \phi}{e}$.
Multiplying by $e$: $2h\nu - 2\phi = 1.5h\nu - \phi$.
Rearranging the terms: $2h\nu - 1.5h\nu = 2\phi - \phi$.
Therefore,$\phi = 0.5h\nu = \frac{h\nu}{2}$.

(A) The photosensitivity of a metal refers to its ability to emit photoelectrons when light of a suitable frequency is incident on it. A metal with a low work function $(\Phi_0)$ requires less energy to eject an electron, making it more photosensitive. Thus, the Assertion is correct.
The work function is defined as $\Phi_0 = hf_0$, where $h$ is Planck's constant and $f_0$ is the threshold frequency. This is a standard physical relationship for the photoelectric effect. Thus, the Reason is also correct.
Since the work function is defined by the threshold frequency $(f_0)$, and a smaller $f_0$ (and thus smaller $\Phi_0$) makes it easier to eject electrons, the Reason correctly explains why a small work function leads to high photosensitivity. Therefore, the correct option is $A$.

(C) According to Einstein's photoelectric equation,$K_{max} = eV_0 = h\nu - \phi$,where $\nu$ is the frequency of incident light and $\phi$ is the work function of the metal.
Since the frequency of $X-$ rays is significantly higher than that of ultraviolet $(U.V.)$ light,the maximum kinetic energy $(K_{max})$ and the stopping potential $(V_0)$ will both increase when $X-$ rays are used.
Therefore,the Assertion is correct.
The Reason states that photoelectrons are emitted with a range of speeds due to a range of frequencies in the incident light. This is incorrect because photoelectric emission occurs even with monochromatic light (a single frequency),and the range of speeds arises due to energy losses of electrons within the metal before emission,not because of a range of incident frequencies.
Thus,the Assertion is correct but the Reason is incorrect.

(B) The Assertion is correct because even for a monochromatic source, electrons emitted from deeper layers of the metal lose energy due to collisions before escaping, resulting in a range of kinetic energies from $0$ to $K_{max}$.
The Reason is also correct because, according to Einstein's photoelectric equation $K_{max} = h\nu - \Phi$, if the incident radiation has different wavelengths (and thus different frequencies $\nu$), the energy acquired by the electrons will vary.
However, the Reason is not the correct explanation for the Assertion because the primary reason for the variation in kinetic energy (even for monochromatic light) is the energy loss during the escape of electrons from different depths within the metal.

(D) The work function $\phi$ is related to the threshold wavelength $\lambda_0$ by the equation $\phi = \frac{hc}{\lambda_0}$.
Given $\phi = 4.0 \ eV$ and using the approximation $hc \approx 1240 \ eV \cdot nm$:
$\lambda_0 = \frac{hc}{\phi} = \frac{1240 \ eV \cdot nm}{4.0 \ eV}$.
$\lambda_0 = 310 \ nm$.
Therefore,the longest wavelength of light that can cause photoemission is $310 \ nm$.

(C) The intensity of the radiation is $I = 6.4 \times 10^{-5} \; W/cm^{2}$ and the area is $A = 1 \; cm^{2}$.
The power incident on the surface is $P = I \times A = 6.4 \times 10^{-5} \; W$.
The energy of a single photon is $E_{ph} = \frac{hc}{\lambda} = \frac{1240 \; eV \cdot nm}{310 \; nm} = 4 \; eV$.
Converting this energy to Joules: $E_{ph} = 4 \times 1.6 \times 10^{-19} \; J = 6.4 \times 10^{-19} \; J$.
The number of photons incident per second $(n)$ is given by $n = \frac{P}{E_{ph}} = \frac{6.4 \times 10^{-5} \; J/s}{6.4 \times 10^{-19} \; J} = 10^{14} \; \text{photons/s}$.
Given that one in $10^{3}$ photons ejects an electron, the number of electrons emitted per second is $N_e = n \times 10^{-3} = 10^{14} \times 10^{-3} = 10^{11}$.
Comparing this with $10^{x}$, we get $x = 11$.

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy.
Given $\lambda_{B} = 2 \lambda_{A}$,we have $\frac{h}{\sqrt{2m T_{B}}} = 2 \frac{h}{\sqrt{2m T_{A}}}$.
Squaring both sides,we get $\frac{1}{T_{B}} = \frac{4}{T_{A}}$,which implies $T_{A} = 4 T_{B}$.
We are given $T_{B} = T_{A} - 1.5$. Substituting $T_{A} = 4 T_{B}$,we get $T_{B} = 4 T_{B} - 1.5$.
This simplifies to $3 T_{B} = 1.5$,so $T_{B} = 0.5 \; eV$.
Using Einstein's photoelectric equation for metal $B$: $K_{max} = E - \phi_{B}$,where $E = 4.5 \; eV$.
$0.5 = 4.5 - \phi_{B}$.
Therefore,$\phi_{B} = 4.5 - 0.5 = 4.0 \; eV$.

(A) Let the work function be $\phi$.
According to Einstein's photoelectric equation,the maximum kinetic energy $KE_{\max}$ is given by:
$KE_{\max} = \frac{hc}{\lambda} - \phi$.
When an electron with charge $q$ and mass $m$ enters a magnetic field $B$ perpendicular to its velocity,it follows a circular path of radius $R = \frac{\sqrt{2m KE_{\max}}}{qB}$.
Squaring both sides,we get $R^2 = \frac{2m KE_{\max}}{q^2 B^2}$,which implies $KE_{\max} = \frac{R^2 q^2 B^2}{2m}$.
Substituting the values: $h = 6.63 \times 10^{-34} \; J \cdot s$,$c = 3 \times 10^8 \; m/s$,$\lambda = 6561 \times 10^{-10} \; m$,$R = 10 \times 10^{-3} \; m$,$q = 1.6 \times 10^{-19} \; C$,$m = 9.1 \times 10^{-31} \; kg$,$B = 3 \times 10^{-4} \; T$.
Energy of incident photon $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6561 \times 10^{-10}} \approx 3.03 \times 10^{-19} \; J = 1.89 \; eV$.
Kinetic energy $KE_{\max} = \frac{(10^{-2})^2 \times (1.6 \times 10^{-19})^2 \times (3 \times 10^{-4})^2}{2 \times 9.1 \times 10^{-31}} \approx 1.26 \times 10^{-20} \; J = 0.079 \; eV$.
Thus,$\phi = E - KE_{\max} = 1.89 \; eV - 0.079 \; eV \approx 1.81 \; eV$.
Wait,re-evaluating the calculation: $E = 1.89 \; eV$. $KE_{\max} = 0.079 \; eV$. $\phi = 1.81 \; eV$.
Given the options,$1.8 \; eV$ is the closest value.

(N/A) For the cut-off or threshold frequency,the energy $h \nu_{0}$ of the incident radiation must be equal to the work function $\phi_{0}$,so that
$\nu_{0} = \frac{\phi_{0}}{h} = \frac{2.14 \ eV}{6.63 \times 10^{-34} \ J \cdot s}$
$= \frac{2.14 \times 1.6 \times 10^{-19} \ J}{6.63 \times 10^{-34} \ J \cdot s} = 5.16 \times 10^{14} \ Hz$
Thus,for frequencies less than this threshold frequency,no photoelectrons are ejected.
$(b)$ Photocurrent reduces to zero when the maximum kinetic energy of the emitted photoelectrons equals the potential energy $e V_{0}$ provided by the retarding potential $V_{0}$. Einstein's photoelectric equation is
$e V_{0} = h \nu - \phi_{0} = \frac{h c}{\lambda} - \phi_{0}$
$\lambda = \frac{h c}{e V_{0} + \phi_{0}}$
$= \frac{(6.63 \times 10^{-34} \ J \cdot s) \times (3 \times 10^{8} \ m/s)}{(0.60 \ eV + 2.14 \ eV) \times 1.6 \times 10^{-19} \ J/eV}$
$= \frac{19.89 \times 10^{-26} \ J \cdot m}{4.384 \times 10^{-19} \ J} \approx 454 \ nm$

(N/A) The energy of an incident photon is given by $E = h\nu = hc / \lambda$.
Using $h = 6.63 \times 10^{-34} \; J s$ and $c = 3 \times 10^8 \; m/s$,we get $hc = 1.989 \times 10^{-25} \; J m$.
$(i)$ For violet light,$\lambda_1 = 390 \; nm = 390 \times 10^{-9} \; m$:
$E_1 = (1.989 \times 10^{-25}) / (390 \times 10^{-9}) \approx 5.10 \times 10^{-19} \; J = 3.19 \; eV$.
$(ii)$ For yellow-green light,$\lambda_2 = 550 \; nm = 550 \times 10^{-9} \; m$:
$E_2 = (1.989 \times 10^{-25}) / (550 \times 10^{-9}) \approx 3.62 \times 10^{-19} \; J = 2.26 \; eV$.
$(iii)$ For red light,$\lambda_3 = 760 \; nm = 760 \times 10^{-9} \; m$:
$E_3 = (1.989 \times 10^{-25}) / (760 \times 10^{-9}) \approx 2.62 \times 10^{-19} \; J = 1.64 \; eV$.
$(b)$ $A$ photoelectric device operates if the incident photon energy $E \geq \phi_0$.
For violet light $(3.19 \; eV)$,it can operate with $Cs$ $(2.14 \; eV)$,$K$ $(2.30 \; eV)$,and $Na$ $(2.75 \; eV)$.
For yellow-green light $(2.26 \; eV)$,it can operate with $Cs$ $(2.14 \; eV)$.
For red light $(1.64 \; eV)$,no material listed has a work function low enough to operate.

(N/A) Given: Work function $\phi_{0} = 2.14 \; eV = 2.14 \times 1.6 \times 10^{-19} \; J = 3.424 \times 10^{-19} \; J$. Frequency $\nu = 6 \times 10^{14} \; Hz$. Planck's constant $h = 6.63 \times 10^{-34} \; J \cdot s$. Mass of electron $m = 9.11 \times 10^{-31} \; kg$.
$(a)$ Using Einstein's photoelectric equation,$K_{\max} = h\nu - \phi_{0}$.
$h\nu = (6.63 \times 10^{-34}) \times (6 \times 10^{14}) = 3.978 \times 10^{-19} \; J$.
$K_{\max} = 3.978 \times 10^{-19} - 3.424 \times 10^{-19} = 0.554 \times 10^{-19} \; J$.
In $eV$,$K_{\max} = (0.554 \times 10^{-19}) / (1.6 \times 10^{-19}) \approx 0.346 \; eV$.
$(b)$ Stopping potential $V_{0}$ is given by $eV_{0} = K_{\max}$.
$V_{0} = (0.554 \times 10^{-19} \; J) / (1.6 \times 10^{-19} \; C) = 0.346 \; V$.
$(c)$ Maximum speed $v_{\max}$ is given by $K_{\max} = \frac{1}{2}mv_{\max}^{2}$.
$v_{\max} = \sqrt{\frac{2K_{\max}}{m}} = \sqrt{\frac{2 \times 0.554 \times 10^{-19}}{9.11 \times 10^{-31}}} = \sqrt{0.1216 \times 10^{12}} \approx 3.49 \times 10^{5} \; m/s = 349 \; km/s$.

(B) The maximum kinetic energy $(K_{max})$ of emitted photoelectrons is related to the cut-off voltage (stopping potential,$V_o$) by the equation:
$K_{max} = e V_o$
Given that the cut-off voltage $V_o = 1.5 \; V$ and the elementary charge $e = 1.6 \times 10^{-19} \; C$,we substitute these values:
$K_{max} = (1.6 \times 10^{-19} \; C) \times (1.5 \; V)$
$K_{max} = 2.4 \times 10^{-19} \; J$
Therefore,the maximum kinetic energy is $2.4 \times 10^{-19} \; J$.

(B) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by $V_0 = (h/e)\nu - \phi/e$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m = h/e$.
Given,slope $m = 4.12 \times 10^{-15} \; V \cdot s$.
We know the charge of an electron $e = 1.6 \times 10^{-19} \; C$.
Therefore,$h = m \times e = (4.12 \times 10^{-15}) \times (1.6 \times 10^{-19}) \; J \cdot s$.
$h = 6.592 \times 10^{-34} \; J \cdot s$.

(C) Given:
Threshold frequency $\nu_{0} = 3.3 \times 10^{14} \;Hz$
Incident frequency $\nu = 8.2 \times 10^{14} \;Hz$
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = h(\nu - \nu_{0})$.
The cutoff voltage (stopping potential) $V_{0}$ in $eV$ is numerically equal to the maximum kinetic energy in $eV$,where $eV_{0} = h(\nu - \nu_{0})$.
Using Planck's constant $h \approx 4.135 \times 10^{-15} \;eV \cdot s$:
$V_{0} = \frac{h(\nu - \nu_{0})}{e} = 4.135 \times 10^{-15} \times (8.2 - 3.3) \times 10^{14}$
$V_{0} = 4.135 \times 10^{-15} \times 4.9 \times 10^{14}$
$V_{0} = 4.135 \times 0.49 \approx 2.026 \;V$.
Rounding to two decimal places,the cutoff voltage is $2.03 \;V$.

(NO) The work function $\phi_{0} = 4.2 \; eV$.
To find the threshold frequency $\nu_{0}$,we use $\phi_{0} = h\nu_{0}$,where $h = 6.63 \times 10^{-34} \; J \cdot s$.
$\nu_{0} = \frac{\phi_{0}}{h} = \frac{4.2 \times 1.6 \times 10^{-19} \; J}{6.63 \times 10^{-34} \; J \cdot s} \approx 1.01 \times 10^{15} \; Hz$.
The frequency of the incident radiation $\nu$ is given by $\nu = \frac{c}{\lambda}$.
$\nu = \frac{3 \times 10^{8} \; m/s}{330 \times 10^{-9} \; m} \approx 0.91 \times 10^{15} \; Hz$.
Since the incident frequency $\nu \approx 0.91 \times 10^{15} \; Hz$ is less than the threshold frequency $\nu_{0} \approx 1.01 \times 10^{15} \; Hz$,photoelectric emission will not occur.

(A) Given: Frequency of incident light $\nu = 7.21 \times 10^{14} \; Hz$,maximum speed of electrons $v = 6.0 \times 10^{5} \; m/s$,mass of electron $m = 9.11 \times 10^{-31} \; kg$,and Planck's constant $h = 6.626 \times 10^{-34} \; J \cdot s$.
According to Einstein's photoelectric equation: $K_{max} = h\nu - h\nu_0$,where $K_{max} = \frac{1}{2}mv^2$.
Substituting the values: $\frac{1}{2} \times (9.11 \times 10^{-31}) \times (6.0 \times 10^5)^2 = (6.626 \times 10^{-34}) \times (7.21 \times 10^{14} - \nu_0)$.
$1.6398 \times 10^{-19} = (6.626 \times 10^{-34}) \times (7.21 \times 10^{14} - \nu_0)$.
$2.4748 \times 10^{14} = 7.21 \times 10^{14} - \nu_0$.
$\nu_0 = 7.21 \times 10^{14} - 2.4748 \times 10^{14} = 4.7352 \times 10^{14} \; Hz$.
Thus,the threshold frequency is approximately $4.73 \times 10^{14} \; Hz$.

(B) Given: Wavelength $\lambda = 488 \;nm = 488 \times 10^{-9} \;m$, Stopping potential $V_o = 0.38 \;V$.
The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $h = 6.63 \times 10^{-34} \;J \cdot s$ and $c = 3 \times 10^8 \;m/s$:
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}} \approx 4.076 \times 10^{-19} \;J$.
Converting to $eV$: $E = \frac{4.076 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.547 \;eV$.
According to Einstein's photoelectric equation: $K_{max} = E - \phi_o$, where $K_{max} = eV_o$.
Thus, $\phi_o = E - eV_o$.
$\phi_o = 2.547 \;eV - 0.38 \;eV = 2.167 \;eV$.
Rounding to the nearest option, the work function is $2.16 \;eV$.

(N/A) Given:
Wavelength of ultraviolet light, $\lambda = 2271 \,\mathring{A} = 2271 \times 10^{-10} \, m$
Stopping potential, $V_{0} = 1.3 \, V$
Planck's constant, $h = 6.63 \times 10^{-34} \, J \cdot s$
Speed of light, $c = 3 \times 10^{8} \, m/s$
Charge of an electron, $e = 1.6 \times 10^{-19} \, C$
Using Einstein's photoelectric equation: $\phi_{0} = \frac{hc}{\lambda} - eV_{0}$
$\phi_{0} = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^{8})}{2271 \times 10^{-10}} - (1.6 \times 10^{-19} \times 1.3)$
$\phi_{0} = 8.758 \times 10^{-19} \, J - 2.08 \times 10^{-19} \, J = 6.678 \times 10^{-19} \, J$
In electron volts: $\phi_{0} = \frac{6.678 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 4.17 \, eV$
For red light: $\lambda_{r} = 6328 \,\mathring{A} = 6328 \times 10^{-10} \, m$
Energy of red light photon, $E_{r} = \frac{hc}{\lambda_{r}} = \frac{1.989 \times 10^{-25}}{6328 \times 10^{-10}} \approx 3.14 \times 10^{-19} \, J \approx 1.96 \, eV$
Since $E_{r} < \phi_{0}$ $(1.96 \, eV < 4.17 \, eV)$, the red light will not cause photoelectric emission, regardless of its high intensity.

(B) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
Using the photoelectric equation: $eV_0 = \frac{hc}{\lambda} - \phi_0$,where $\phi_0$ is the work function.
For the neon lamp: $\lambda_1 = 640.2 \;nm$,$V_{01} = 0.54 \;V$.
$\phi_0 = \frac{hc}{\lambda_1} - eV_{01} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{640.2 \times 10^{-9}} - 0.54 \;eV = 1.94 \;eV - 0.54 \;eV = 1.40 \;eV$.
For the iron source: $\lambda_2 = 427.2 \;nm$.
$eV_{02} = \frac{hc}{\lambda_2} - \phi_0 = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{427.2 \times 10^{-9}} - 1.40 \;eV$.
$eV_{02} = 2.90 \;eV - 1.40 \;eV = 1.50 \;eV$.
Thus,the new stopping voltage is $1.5 \;V$.

(N/A) Einstein's photoelectric equation is given by:
$e V_0 = h \nu - \phi_0$
$V_0 = \frac{h}{e} \nu - \frac{\phi_0}{e} \dots (i)$
Where $V_0$ is the stopping potential,$h$ is Planck's constant,$e$ is the charge of an electron,$\nu$ is the frequency of radiation,and $\phi_0$ is the work function.
Using $\nu = \frac{c}{\lambda}$ (where $c = 3 \times 10^8 \,m/s$),we calculate the frequencies:
$\nu_1 = 8.219 \times 10^{14} \,Hz, \nu_2 = 7.412 \times 10^{14} \,Hz, \nu_3 = 6.884 \times 10^{14} \,Hz, \nu_4 = 5.493 \times 10^{14} \,Hz, \nu_5 = 4.343 \times 10^{14} \,Hz$

Frequency $(\times 10^{14} \,Hz)$	$8.219$	$7.412$	$6.884$	$5.493$	$4.343$
Stopping potential $V_0$ $(V)$	$1.28$	$0.95$	$0.74$	$0.16$	$0$

The graph of $V_0$ versus $\nu$ is a straight line. The threshold frequency $\nu_0$ is the intercept on the $\nu$-axis,which is $5 \times 10^{14} \,Hz$.
The slope of the line is $\frac{h}{e} = \frac{1.28 - 0.16}{(8.219 - 5.493) \times 10^{14}} = \frac{1.12}{2.726 \times 10^{14}} \approx 4.108 \times 10^{-15} \,V \cdot s$.
$h = (4.108 \times 10^{-15}) \times (1.6 \times 10^{-19}) \approx 6.573 \times 10^{-34} \,J \cdot s$.
Work function $\phi_0 = h \nu_0 = (6.573 \times 10^{-34}) \times (5 \times 10^{14}) = 3.286 \times 10^{-19} \,J = 2.054 \,eV$.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Given $\lambda = 3300\,\mathring{A} = 3300 \times 10^{-10}\,m$,$h = 6.63 \times 10^{-34}\,Js$,and $c = 3 \times 10^8\,m/s$.
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3300 \times 10^{-10}}\,J = 6.027 \times 10^{-19}\,J$.
Converting to electron-volts: $E = \frac{6.027 \times 10^{-19}}{1.6 \times 10^{-19}}\,eV \approx 3.77\,eV$.
Photoelectric emission occurs only if the incident photon energy $E$ is greater than the work function $\Phi$ of the metal.
Here,$E = 3.77\,eV$. Comparing this with the given work functions:
$Na (2.75\,eV) < 3.77\,eV$ (Emission occurs)
$K (2.30\,eV) < 3.77\,eV$ (Emission occurs)
$Mo (4.17\,eV) > 3.77\,eV$ (No emission)
$Ni (5.15\,eV) > 3.77\,eV$ (No emission)
Thus,$Mo$ and $Ni$ will not show photoelectric emission.
If the laser is moved to $50\;cm$,the intensity of the radiation increases,but the energy of individual photons remains $3.77\,eV$. Since the energy is still less than the work functions of $Mo$ and $Ni$,there will still be no photoelectric emission for these metals.

(N/A) The threshold frequency $(\nu_0)$ is the minimum frequency of incident radiation required to eject electrons from a metal surface.
Each metal has a unique work function $(\Phi = h\nu_0)$, which determines its specific threshold frequency.
If the frequency of incident radiation $(\nu)$ is less than the threshold frequency $(\nu < \nu_0)$, the energy of the incident photons is insufficient to overcome the work function of the metal, and no photoelectric emission occurs, regardless of the intensity of the light.
If $\nu > \nu_0$, the energy of the incident photons is greater than the work function, and photoelectric emission occurs instantaneously (within $10^{-9} \, s$ or less), even if the intensity of the radiation is very low.

(D) According to Einstein's photoelectric equation, $K_{max} = h\nu - \Phi_0$, where $K_{max}$ is the maximum kinetic energy of the photoelectron, $h$ is Planck's constant, $\nu$ is the frequency of incident light and $\Phi_0$ is the work function. The work function $\Phi_0$ depends on the type of metal surface. Therefore, the energy of the photoelectron depends on the frequency of incident light ($\nu$) and the type of metal surface ($\Phi_0$). Thus, the correct option is $(D)$.

(N/A) According to the wave theory of light,the energy of a wave is proportional to the square of its amplitude (intensity) and is independent of its frequency.
$1$. Wave theory suggests that the energy delivered to the metal surface depends on the intensity of the light,not its frequency.
$2$. Therefore,if the intensity is kept constant,the wave theory predicts that the kinetic energy of the emitted electrons should remain constant regardless of the frequency of the incident light.
$3$. However,experimental observations show that the maximum kinetic energy of photoelectrons increases linearly with the frequency of incident light.
$4$. This discrepancy arises because light interacts with matter as discrete packets of energy called photons,where the energy of each photon is given by $E = h\nu$. Since the wave theory treats light as a continuous wave,it fails to account for this frequency-dependent energy transfer.

(N/A) In $1905$,Einstein proposed a new theory to explain the photoelectric effect.
According to this theory,the emission and absorption of energy (called a photon) take place in discrete units. These units are called quanta of energy of radiation.
Each quantum (photon) has energy $E = h\nu$,where $\nu$ is the frequency of radiation. An electron on the surface will absorb this energy $h\nu$.
If the energy absorbed by an electron on the surface is greater than the minimum energy required (work function $\phi_{0}$),then the electron will be emitted with maximum kinetic energy $K_{\max}$.
Let the energy of the incident radiation be $h\nu$,the maximum kinetic energy of the electron be $K_{\max}$,and the work function of the metal be $\phi_{0}$. Then,according to the law of conservation of energy:
$h\nu = K_{\max} + \phi_{0}$
Therefore,$K_{\max} = h\nu - \phi_{0} \quad \dots (1)$
More tightly bound electrons will emerge with kinetic energy less than the maximum value. With an increase in the intensity of light,the number of electrons emitted per second increases. However,the maximum kinetic energy of the emitted electron is determined solely by the energy of the photon.
In Einstein's equation,the maximum kinetic energy is $K_{\max} = e V_{0}$,where $V_{0}$ is the stopping potential. Substituting this into equation $(1)$:
$e V_{0} = h\nu - \phi_{0} \quad \dots (2)$
Rearranging this gives $V_{0} = \left(\frac{h}{e}\right)\nu - \frac{\phi_{0}}{e}$.
The graph of stopping potential $V_{0}$ versus frequency $\nu$ is a straight line. The slope of the $V_{0}-\nu$ graph is $\frac{h}{e}$,which is a universal constant and does not depend on the type of material used.

(N/A) Einstein's photoelectric equation is given by:
$\frac{1}{2} m v_{\max }^{2} = h \nu - \phi_{0}$
Since the maximum kinetic energy is related to the stopping potential $V_{0}$ by the relation:
$\frac{1}{2} m v_{\max }^{2} = e V_{0}$
Substituting this into the photoelectric equation,we get:
$e V_{0} = h \nu - \phi_{0}$
Rearranging for $V_{0}$:
$V_{0} = \left( \frac{h}{e} \right) \nu - \frac{\phi_{0}}{e}$
This equation is of the form $y = mx + c$,representing a straight line where the slope is $\frac{h}{e}$.
Millikan performed a series of experiments measuring the stopping potential $V_{0}$ for different frequencies $\nu$ of incident radiation. He plotted a graph of $V_{0}$ versus $\nu$,which resulted in a straight line. The slope of this line was found to be $\frac{h}{e}$.
By using the known value of the elementary charge $e$,Millikan calculated the value of Planck's constant $h$ to be approximately $6.626 \times 10^{-34} \text{ Js}$,which matched the previously accepted value.
Thus,while Millikan initially intended to disprove Einstein's theory,his experimental results provided strong evidence for its validity,confirming the photoelectric equation with great precision across various alkali metals.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of a 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.
$1$. Dependence on frequency: The equation shows that $K_{max}$ increases linearly with the frequency $(\nu)$ of the incident radiation.
$2$. Independence: $K_{max}$ does not depend on the intensity of the incident radiation,as intensity only affects the number of photoelectrons emitted per second,not their individual kinetic energies.

(N/A) The kinetic energy $(K_{max})$ of a photoelectron is given by Einstein's photoelectric equation: $K_{max} = h\nu - \Phi_0$,where $h\nu$ is the energy of the incident photon and $\Phi_0$ is the work function of the metal surface.
$1$. By definition,kinetic energy is the energy possessed by an object due to its motion,which is given by the formula $K = \frac{1}{2}mv^2$. Since mass $(m)$ is always positive and the square of velocity $(v^2)$ is always non-negative,kinetic energy can never be negative.
$2$. Physically,if the incident photon energy $(h\nu)$ is less than the work function $(\Phi_0)$,the photon does not have enough energy to eject an electron from the metal surface at all. In this case,no photoelectric emission occurs,and the kinetic energy is not defined as negative; rather,the process simply does not take place.

(N/A) The photoelectric equation is given by $h\nu = \phi_0 + eV_0$,where $h$ is Planck's constant,$\nu$ is the frequency of incident radiation,$\phi_0$ is the work function,and $V_0$ is the stopping potential.
Rearranging for $V_0$,we get $V_0 = (\frac{h}{e})\nu - \frac{\phi_0}{e}$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m$ is $\frac{h}{e}$.
The slope of the $V_0 - \nu$ graph is $\frac{h}{e}$,which is a universal constant.
Since $h$ (Planck's constant) and $e$ (charge of an electron) are universal constants,the slope does not depend on the type of material used for the photosensitive surface.

(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.
Since $K_{max} = eV_0$, where $e$ is the charge of an electron and $V_0$ is the stopping potential, we can write the equation as $eV_0 = h\nu - \Phi_0$.
Rearranging this to the form $y = mx + c$, we get $V_0 = (h/e)\nu - (\Phi_0/e)$.
However, for the $eV_0 - \nu$ graph, the equation is $eV_0 = h\nu - \Phi_0$.
Comparing this with $y = mx + c$, where $y = eV_0$ and $x = \nu$, the slope $m$ is equal to $h$.

(A) $(i)$ Suppose one electron absorbs two photons each having frequency $f$. Out of this total energy $2hf$,it spends $W$ amount of energy to overcome the work function for its emission,and the remaining amount $(2hf - W)$ is possessed by it in the form of kinetic energy after being emitted.
Hence,$K = 2hf - W$.
Therefore,$K_{max} = 2hf - W_{min}$.
Since $W_{min} = \phi_0$,we get $K_{max} = 2hf - \phi_0$.
$(ii)$ If the above assumption were true,then according to the work-energy theorem,taking $K_{max} = V_0 e$ (where $V_0$ is the stopping potential):
$2hf - \phi_0 = V_0 e$
$V_0 = (\frac{2h}{e})f - (\frac{\phi_0}{e})$
This equation represents a straight line $y = mx + c$. If we plot a graph of $V_0$ versus $f$ experimentally,the slope should be $(\frac{2h}{e})$. However,experimental results consistently show the slope to be $(\frac{h}{e})$. Thus,the assumption of two-photon absorption is incorrect,which is why it is not considered in the discussion of stopping potential.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - \phi_0$,where $\phi_0$ is the work function.
For $\lambda_1 = 600 \, nm$,$K_1 = \frac{hc}{600} - \phi_0$ $(1)$
For $\lambda_2 = 400 \, nm$,$K_2 = \frac{hc}{400} - \phi_0$ $(2)$
Given $K_2 = 2K_1$,we substitute $(1)$ into $(2)$:
$\frac{hc}{400} - \phi_0 = 2 \left( \frac{hc}{600} - \phi_0 \right)$
$\frac{hc}{400} - \phi_0 = \frac{hc}{300} - 2\phi_0$
$\phi_0 = hc \left( \frac{1}{300} - \frac{1}{400} \right) \times 10^9$
Using $hc \approx 1240 \, eV \cdot nm$:
$\phi_0 = 1240 \left( \frac{4-3}{1200} \right) = \frac{1240}{1200} \approx 1.03 \, eV$.
Rounding to the nearest option,the work function is $1.0 \, eV$.

(N/A) Zero-point energy is the lowest possible energy that a quantum mechanical physical system may have. Unlike in classical mechanics,quantum systems fluctuate even in their ground state. For a gas of fermions,the energy of the highest occupied state at absolute zero temperature $(0 \ K)$ is called the Fermi energy $(E_F)$.

(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi = eV,$ where $\phi = \frac{hc}{\lambda_0}$ is the work function.
For wavelength $\lambda$: $\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + eV$ ... $(i)$
For wavelength $3\lambda$: $\frac{hc}{3\lambda} = \frac{hc}{\lambda_0} + \frac{eV}{4}$ ... (ii)
Multiply equation (ii) by $4$: $\frac{4hc}{3\lambda} = \frac{4hc}{\lambda_0} + eV$ ... (iii)
Equating the expressions for $eV$ from $(i)$ and (iii):
$\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{4hc}{3\lambda} - \frac{4hc}{\lambda_0}$
Rearranging terms: $\frac{4hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{4hc}{3\lambda} - \frac{hc}{\lambda}$
$\frac{3hc}{\lambda_0} = \frac{hc}{3\lambda}$
$\frac{3}{\lambda_0} = \frac{1}{3\lambda} \implies \lambda_0 = 9\lambda$
Given $\lambda_0 = n\lambda,$ therefore $n = 9.$

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
Let $\lambda_1 = 500 \, nm$ and $\lambda_2 = 200 \, nm$.
Let $K_1$ and $K_2$ be the maximum kinetic energies corresponding to $\lambda_1$ and $\lambda_2$ respectively.
Given $K_2 = 3K_1$.
Using $hc = 1240 \, eV \cdot nm$:
$K_1 = \frac{1240}{500} - \phi = 2.48 - \phi$
$K_2 = \frac{1240}{200} - \phi = 6.2 - \phi$
Substituting into $K_2 = 3K_1$:
$6.2 - \phi = 3(2.48 - \phi)$
$6.2 - \phi = 7.44 - 3\phi$
$2\phi = 7.44 - 6.2$
$2\phi = 1.24$
$\phi = 0.62 \, eV$.
The closest value is $0.61 \, eV$.

(C) According to Einstein's photoelectric equation,the stopping potential $V$ is given by:
$eV = \frac{hc}{\lambda} - \phi$
Rearranging this for $V$:
$V = \left(\frac{hc}{e}\right)\left(\frac{1}{\lambda}\right) - \frac{\phi}{e}$
Comparing this with the equation of a straight line $y = mx + c$,we have:
Slope $m = \frac{hc}{e}$ and intercept $c = -\frac{\phi}{e}$.
Here,$h$ is Planck's constant,$c$ is the speed of light,$e$ is the charge of an electron,and $\phi$ is the work function of the metal surface.
None of these parameters $(h, c, e, \phi)$ depend on the intensity of the incident radiation. The intensity of radiation only affects the number of photoelectrons emitted per unit time (photoelectric current),not their maximum kinetic energy or the stopping potential.
Therefore,changing the intensity of the incident radiation does not affect the slope or the intercept of the graph. Thus,the graph remains unchanged.

(D) The graph shows the variation of stopping potential $V_s$ with frequency $f$ of incident light.
According to Einstein's photoelectric equation: $hf = \phi + eV_s$,where $\phi$ is the work function.
At the threshold frequency $f_0$,the stopping potential $V_s = 0$.
From the graph,the intercept on the frequency axis (where $V_s = 0$) is $f_0 = 5 \times 10^{14} \, Hz$.
The work function $\phi$ is given by $\phi = hf_0$.
Substituting the values: $\phi = (6.62 \times 10^{-34} \, J \cdot s) \times (5 \times 10^{14} \, Hz) = 33.1 \times 10^{-20} \, J$.
To convert this energy into electron-volts $(eV)$,divide by the charge of an electron $(e = 1.6 \times 10^{-19} \, C)$:
$\phi = \frac{33.1 \times 10^{-20}}{1.6 \times 10^{-19}} \, eV = 2.06875 \, eV \approx 2.07 \, eV$.
Given the options,the closest value is $2.10 \, eV$.