Dual Nature of Radiation

The power $(P)$ of the laser source is, $P=6 \mathrm{~mW}=6 \times 10^{-3} \mathrm{~J} / \mathrm{s}$

The wavelength ( $\lambda$ ) of the emitted photon is $\lambda=663 \mathrm{~nm}=663 \times 10^{-9} \mathrm{~m}$

Planck's constant $\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s}$

Speed of light $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}$

The energy of a single photon is given by :

$ \mathrm{E}=\frac{\mathrm{hc}}{\lambda} $

Total Power is the number of photons per second multiplied by the energy of one photon:

$ \mathrm{P}=\mathrm{n} \times \mathrm{E} \Rightarrow \mathrm{n}=\frac{\mathrm{P}}{\mathrm{E}}=\frac{\mathrm{P} \lambda}{\mathrm{hc}} $

Substituting the given values into the equation :

$n=\frac{\left(6 \times 10^{-3}\right) \times\left(663 \times 10^{-9}\right)}{\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}$

$\Rightarrow $ $\mathrm{n}=\frac{600}{3} \times \frac{10^{-12}}{10^{-26}}$

$\Rightarrow $ $\mathrm{n}=200 \times 10^{14}$

$\Rightarrow $ $ \mathrm{n}=2 \times 10^{16} $

Therefore, the laser source emits $2 \times 10^{16}$ photons every second. Hence, the correct option is (D).

When a light of wavelength ( $\lambda$ ) is incident on a metal of work function $\phi$ then the maximum kinetic energy of the photo-electron is given by Einstein's photoelectric equation:

$ \mathrm{K}_{\max }=\mathrm{E}-\phi \Rightarrow \mathrm{E}=\phi+\mathrm{K}_{\max } $

$ \Rightarrow \frac{\mathrm{hc}}{\lambda}=\phi+\mathrm{eV}_{\mathrm{s}} $

where $\mathrm{V}_{\mathrm{s}}$ is the stopping potential and $\phi$ is the work function of the metal.

Case 1 : Wavelength is $\lambda$, stopping potential is 3.2 V .

$ \frac{\mathrm{hc}}{\lambda}=\phi+3.2 \mathrm{e} $.....(i)

Case 2 : Wavelength is $2 \lambda$, stopping potential is 0.7 V .

$ \frac{\mathrm{hc}}{2 \lambda}=\phi+0.7 \mathrm{e} \ldots (ii)$

Subtracting equation (ii) from equation (i) :

$ \left(\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{2 \lambda}\right)=(3.2 \mathrm{e}-0.7 \mathrm{e}) $

$\Rightarrow $ $\frac{\mathrm{hc}}{2 \lambda}=2.5 \mathrm{e}$

$\Rightarrow $ $ \lambda=\frac{\mathrm{hc}}{5 \mathrm{e}} $

Substituting the given values :

$ \lambda=\frac{\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{5 \times\left(1.6 \times 10^{-19}\right)} $

$\Rightarrow $ $ \lambda=\frac{19.89 \times 10^{-26}}{8 \times 10^{-19}} \mathrm{~m} \approx 2.5 \times 10^{-7} \mathrm{~m} $

Therefore, the initial wavelength of light used was $2.5 \times 10^{-7} \mathrm{~m}$.

Hence, the correct option is (B).

For a gas molecule, the average kinetic energy $(\mathrm{K})$ is given by kinetic theory:

$ \mathrm{K}=\frac{3}{2} \mathrm{k}_{\mathrm{B}} \mathrm{~T} $

where $\mathrm{k}_{\mathrm{B}}$ is the Boltzmann constant and T is the absolute temperature.

The relationship between momentum ( p ) and kinetic energy ( K ) is:

$ \mathrm{p}=\sqrt{2 \mathrm{mK}} $

Substituting K:

$ p=\sqrt{2 m\left(\frac{3}{2} k_B T\right)}=\sqrt{3{m k_B} \mathrm{~T}} \ldots (i)$

The de Broglie wavelength is defined as:

$ \lambda=\frac{\mathrm{h}}{\mathrm{p}} \ldots $

From both the relations,

$ \lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mk}_{\mathrm{B}} \mathrm{~T}}} $

The given temperature is $27^{\circ} \mathrm{C}$.

$ \mathrm{T}=27+273=300 \mathrm{~K} $

The given mass is $5.31 \times 10^{-26} \mathrm{~kg}$.

So, the de Broglie wavelength is,

$ \lambda=\frac{6.63 \times 10^{-34}}{\sqrt{3 \times 5.31 \times 10^{-26}}}=\frac{6.63 \times 10^{-34}}{\sqrt{6595.02 \times 10^{-49}}}=\frac{6.63 \times 10^{-34}}{\sqrt{6.595 \times 10^{-46}}}=\frac{6.63 \times 10^{-34}}{2.568 \times 10^{-23}} $

$\Rightarrow $ $\lambda \approx 2.58 \times 10^{-11} \mathrm{~m}=25.8 \times 10^{-12} \mathrm{~m}$

$\Rightarrow \mathrm{x} \times 10^{-12} \mathrm{~m}=25.8 \times 10^{-12} \mathrm{~m}$

$ \Rightarrow x \approx 26 $

Therefore, the value of x is 26 . Hence, the correct option is (D).

The energy of the incident photon (E) is used to overcome the work function ( $\Phi$ ) of the metal, and the remaining energy becomes the maximum kinetic energy ( $\mathrm{K}_{\max }$ ) of the emitted photoelectron:

$ \mathrm{K}_{\max }=\mathrm{E}-\Phi $

It is given that the produced photoelectrons have exactly zero kinetic energy ( $K_{\max }=0$ ). This means the energy of the incident light is exactly equal to the work function of the metal (this corresponds to the threshold frequency).

$ \begin{aligned} & 0=E-\Phi \\\\ & \Rightarrow E=\Phi \ldots \text { (i) } \end{aligned} $

The energy of a photon is given by $E=h f$, where $h$ is Planck's constant and $f$ is the linear frequency.

Using the relationship between linear frequency (f) and angular frequency ( $\omega$ ):

$ \omega=2 \pi f \Rightarrow f=\frac{\omega}{2 \pi} \ldots $

So, the expression for the energy of a photon is,

$ \mathrm{E}=\mathrm{h}\left(\frac{\omega}{2 \pi}\right)=\frac{\mathrm{h} \omega}{2 \pi} \ldots $

From relations (i) and (iii)

$ \frac{h \omega}{2 \pi}=\Phi $

$ \Rightarrow \omega=\frac{2 \pi \Phi}{h} $

Work function of metal is, $\Phi=110 \times 10^{-20} \mathrm{~J}$

The Planck's constant is, $\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s}$

Putting the values,

$ \begin{gathered} \omega=\frac{2 \times 3.14 \times 110 \times 10^{-20}}{6.63 \times 10^{-34}} \mathrm{rad} / \mathrm{s} \\ \omega=\frac{691.1 \times 10^{-20}}{6.63 \times 10^{-34}} \\ \omega \approx 1.04 \times 10^{14} \mathrm{rad} / \mathrm{s} \end{gathered} $

Therefore, the angular frequency of the incident light is $1.04 \times 10^{14} \mathrm{rad} / \mathrm{s}$. Hence, the correct option is (D).

According to Einstein's photoelectric effect, the energy of an incident photon (E) on a metal of work function $(\phi)$ then the maximum kinetic energy $\left(\mathrm{K}_{\max }\right)$ to the ejected electron is,

$ \mathrm{K}_{\max }=\mathrm{E}-\phi $

The energy of a photon is given by :

$ \mathrm{E}=\mathrm{hf}=\frac{\mathrm{h} \omega}{2 \pi} $

Where :

• h is Planck's constant.

• $\omega$ is the angular frequency of the light wave.

The given electric field is $\mathrm{E}=60\left[\sin \left(3 \times 10^{15}\right) \mathrm{t}+\sin \left(12 \times 10^{15}\right) \mathrm{t}\right]$

The given electric field equation contains two sinusoidal terms with angular frequencies are $\omega_1=3 \times 10^{15} \mathrm{rad} / \mathrm{s}$ and $\omega_2=12 \times 10^{15} \mathrm{rad} / \mathrm{s}$

Since $\mathrm{K}_{\max }$ depends on the maximum photon energy, we must use the higher angular frequency, $\omega_{\max }=12 \times 10^{15} \mathrm{rad} / \mathrm{s}$, because higher frequency corresponds to higher energy per photon.

The energy in Joules is $E=\frac{h \omega}{2 \pi}$.

$ \mathrm{E}(\text { in } \mathrm{eV})=\frac{\mathrm{h} \cdot \omega_{\max }}{2 \pi \cdot \mathrm{e}} $

Substituting the given values :

$ E=\frac{\left(6.6 \times 10^{-34}\right) \times\left(12 \times 10^{15}\right)}{2 \times 3.14 \times\left(1.6 \times 10^{-19}\right)} $

$\Rightarrow $ $E=\frac{79.2 \times 10^{-19}}{10.048 \times 10^{-19}}$

$\Rightarrow $ $ \mathrm{E} \approx 7.88 \mathrm{eV} $

So, the maximum kinetic energy of ejected electron is,

$ \mathrm{K}_{\max }=\mathrm{E}-\phi $

$\Rightarrow $ $\mathrm{K}_{\max }=7.88 \mathrm{eV}-2.8 \mathrm{eV}$

$\Rightarrow $ $\mathrm{K}_{\max }=5.08 \mathrm{eV}$

$\Rightarrow $ $ \mathrm{K}_{\max } \approx 5.1 \mathrm{eV} $

Therefore, the kinetic energy of electron is approximately 5.1 eV.

Hence, the correct option is (D).

Kinetic Energy (K.E.): The kinetic energy of the emitted electron is given by subtracting the work function (W) of the material from the energy of the incident radiation (E).

$ \text{K.E.} = E - W $

Expressions for Energy:

The energy of the incoming radiation $ E $ is calculated using the formula:

$ E = \frac{hc}{\lambda_i} $

The work function $ W $, which is the minimum energy needed to release an electron, when the radiation wavelength is at its longest $\lambda_0$, is:

$ W = \frac{hc}{\lambda_0} $

De-Broglie Wavelength of the Electron:

The de-Broglie wavelength $\lambda_e$ of the emitted electron is given by:

$ \lambda_e = \frac{h}{\sqrt{2m \cdot \text{K.E.}}} $

Substituting the Values:

The kinetic energy of the emitted electron can be expressed in terms of energies as:

$ \frac{h^2}{2m \lambda_e^2} = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0} $

Final Expression for $\lambda_e$:

Solving for $\lambda_e$, the de-Broglie wavelength of the electron, we have:

$ \lambda_e = \sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}} $

This expression determines the de-Broglie wavelength of the emitted electron based on the wavelengths of the incident and threshold radiation.

Force due to beam assuming complete reflection $\mathrm{F}=\frac{2 \mathrm{P}}{\mathrm{C}}=\frac{2}{\mathrm{C}} \frac{\mathrm{dE}}{\mathrm{dt}} ; \mathrm{P}$ is power

So change in momentum of mirror.

$\mathrm{m}(\mathrm{~V}-0)=\int \mathrm{Fdt}=\frac{2}{\mathrm{C}} \int \mathrm{dE}=\frac{2 \mathrm{E}}{\mathrm{C}}$

Now using work energy theorem ....... (1)

$\begin{aligned} & \mathrm{W}_{\mathrm{g}}=\Delta \mathrm{k} \\ & -\mathrm{mg} \ell(1-\cos \theta)=0-\frac{1}{2} \mathrm{mv}^2 \\ & \mathrm{~g} \ell\left(2 \sin ^2 \frac{\theta}{2}\right)=\frac{\mathrm{v}^2}{2} \end{aligned}$

as $\theta$ is small

$\begin{aligned} & \mathrm{g} \ell 2\left(\frac{\theta}{2}\right)^2=\frac{1}{2} \frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \quad \text{(from eq. (1))}\\ & \mathrm{~g} \ell \theta^2=\frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \\ & \theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{~g} \ell}} \end{aligned}$

In the photoelectric effect, the stopping potential ($V_s$) is determined by the equation:

$ V_s = \frac{hv - \phi}{e} $

where $ h $ is Planck's constant, $ v $ is the frequency of the incident light, $ \phi $ is the work function of the material, and $ e $ is the charge of the electron. This shows that the stopping potential depends on the frequency of the incident light and the work function, not on the intensity of the light.

The intensity of light ($ I $) affects the number of photons hitting the surface per second, given by:

$ I = \frac{\eta hv}{A} $

where $ \eta $ is the efficiency of photoelectron emission and $ A $ is the area. Increasing the intensity increases the number of photons per second ($ n $), which in turn increases the number of emitted electrons. However, the stopping potential remains unaffected by changes in intensity; it is solely influenced by the frequency of the incident light relative to the work function of the material.

To determine which color of visible light will cause the emission of photoelectrons from a metal with a work function of 3 eV, we need to consider the relationship between the photon energy, the work function, and the wavelength of light.

The maximum kinetic energy ($ \text{KE}_{\max} $) of the emitted photoelectrons is given by:

$ \text{KE}_{\max} = \frac{hc}{\lambda} - \phi $

For photoemission to occur, the energy of the incident photons must be greater than the work function ($\phi$):

$ \frac{hc}{\lambda} > \phi $

This implies:

$ \lambda < \frac{hc}{\phi} $

Substitute the values $ h = 4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s} $, $ c = 3 \times 10^{8} \, \text{m/s} $, and $\phi = 3 \, \text{eV}$:

$ \lambda < \frac{1242}{3} \, \text{nm} \approx 414 \, \text{nm} $

Since the wavelength must be less than 414 nm to cause emission, we need light with a shorter wavelength. Among the visible colors, blue light has a wavelength in this range, making it the correct choice. Thus, blue light will cause the emission of photoelectrons with this work function.

First, the intensity $I$ at a distance $r$ from a point light source spreading its energy isotropically is given by

$I = \frac{P}{4\pi r^2},$

where

$P = 450 \, \text{W}$

$r = 2 \, \text{m}.$

So,

$I = \frac{450}{4\pi (2)^2} = \frac{450}{16\pi} \approx \frac{450}{50.27} \approx 8.96 \, \text{W/m}^2.$

For a perfectly reflecting surface, the radiation pressure $p$ is given by

$p = \frac{2I}{c},$

where $c \approx 3 \times 10^8 \, \text{m/s}$ is the speed of light.

Substituting the value of $I$:

$p = \frac{2 \times 8.96}{3 \times 10^8} \approx \frac{17.92}{3 \times 10^8} \, \text{Pa}.$

Calculating the final value:

$p \approx 5.97 \times 10^{-8} \, \text{Pa},$

which is approximately $6 \times 10^{-8} \, \text{Pa}.$

Thus, the correct answer is:

Option D: $6 \times 10^{-8}$ Pascals.

Magnetic field does not work

$\therefore$ Speed will not charge, so De-Broglie wavelength remains same.

To determine the distance between points $A$ and $B$ (where the electron re-enters the metallic plate), we first need to understand the electron's behavior in a magnetic field.

Maximum Kinetic Energy (KE): The energy of the emitted electron can be given by the photoelectric equation:

$ \text{KE}_{\max} = \frac{hc}{\lambda} - \phi $

Here, $\frac{hc}{\lambda}$ is the energy of the incident photons, and $\phi$ is the work function of the metal.

Momentum (p): The momentum of the electron is related to its kinetic energy by the equation:

$ p = \sqrt{2m \cdot \text{KE}_{\max}} = \sqrt{2m\left(\frac{hc}{\lambda} - \phi\right)} $

Path of Electron in Magnetic Field: When an electron moves perpendicularly through a magnetic field, it follows a circular path. The radius $R$ of this path is given by:

$ R = \frac{p}{eB} $

where $e$ is the charge of the electron, and $B$ is the magnetic field strength.

Distance Between $A$ and $B$: Since the electron travels back to the plate, forming a complete semicircle, the distance between points $A$ and $B$ is twice the radius of this circular path:

$ d_{A-B} = 2R = 2\left(\frac{p}{eB}\right) $

Substituting the expression for momentum, we find:

$ d_{A-B} = \frac{2 \sqrt{2m\left(\frac{hc}{\lambda} - \phi\right)}}{eB} = \frac{\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)}}{eB} $

Therefore, the distance between $A$ and $B$ is $\frac{\sqrt{8m(\frac{hc}{\lambda} - \phi)}}{eB}$.

We know, $e{V_0} = hv - \phi $ .... (i)

$ \Rightarrow e{V_0} = K{E_{\max }}$ (where $\phi$ = work function of the metal and $v_0$ = stopping potential)

$ \Rightarrow {V_0} = \left( {{1 \over e}} \right)K{E_{\max }}$

Stopping potential depends solely on the frequency of the incident light $+(v)$ & the work function of the metal, not the intensity.

From eq. (i), we can see,

$v \uparrow \Rightarrow {v_0} \uparrow $

$ \Rightarrow {e \over {\lambda \downarrow }} \Rightarrow {v_0} \uparrow $ (as $v = {c \over \lambda }$)

So, stopping potential increases with decrease in the wavelength of the incident light.

Hence, option 1 is correct.

$\begin{aligned} &\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}} \\ & \lambda^2=\frac{\mathrm{h}^2}{2 \mathrm{~m}}\left(\frac{1}{\mathrm{k}}\right) \\ & \mathrm{Y}=\mathrm{cx}^2 \end{aligned}\\ &\text { Upward facing parabola passing through origin. } \end{aligned}$

Assertion (A) is true because a sufficiently negative potential can repel and stop the emitted electrons from leaving the surface.

Reason (R) is also true because the stopping potential $V_{\text{stop}}$ varies linearly with the frequency $\nu$ of the incident light ($V_{\text{stop}} = \frac{h}{e}\,\nu - \frac{\phi}{e}$).

However, (R) does not directly explain why a negative potential suppresses electron emission; it only shows how much potential is needed for different frequencies.

Therefore, both (A) and (R) are true, but (R) is not the correct explanation of (A).

Solution Explanation

Among the four listed phenomena:

Reflection of light

Wave theory (Huygens’s principle, for example) adequately explains reflection.

Refraction of light

Wave theory also explains refraction via the change of wave speed in different media.

Diffraction of light

Wave theory (again, Huygens–Fresnel principle) explains the bending of waves around obstacles.

Compton effect

This involves inelastic scattering of photons by electrons (where the photon loses energy and is scattered with a longer wavelength).

Wave theory alone cannot explain the discrete energy exchange and wavelength shift; it requires the particle (photon) concept from quantum theory.

Hence, the phenomenon not explained by the (classical) wave theory is the Compton effect.

Answer: (D) Compton effect.

To find the ratio of the de Broglie wavelength of a proton to the wavelength of a photon when they have the same energy, we follow these steps:

Energy Equivalence:

Both the proton and the photon have the same energy, denoted by E.

Energy of the Photon:

The energy of a photon with wavelength λ is given by:

$ E = \frac{hc}{\lambda} $

Kinetic Energy of the Proton:

Since the proton moves at a non-relativistic speed, its kinetic energy can be expressed as:

$ \text{KE}_{\text{proton}} = \frac{1}{2}m_P v^2 = E $

Momentum of the Proton:

The momentum p of the proton is derived from its kinetic energy:

$ p = \sqrt{2m_P E} $

de Broglie Wavelength of the Proton:

The de Broglie wavelength λ_{proton} is:

$ \lambda_{\text{proton}} = \frac{h}{p} = \frac{h}{\sqrt{2m_P E}} $

Ratio of Wavelengths:

The ratio of the de Broglie wavelength of the proton to the wavelength of the photon is:

$ \frac{\lambda_{\text{proton}}}{\lambda_{\text{photon}}} = \left( \frac{h}{\sqrt{2m_P E}} \right) \times \left( \frac{E}{hc} \right) $

Simplifying, we get:

$ \frac{\lambda_{\text{proton}}}{\lambda_{\text{photon}}} = \frac{1}{c} \sqrt{\frac{E}{2m_P}} $

Thus, the ratio of the de Broglie wavelength of the proton to the wavelength of the photon is $\frac{1}{c} \sqrt{\frac{E}{2m_P}}$.

$\begin{aligned} & \mathrm{hv}=\phi+\mathrm{KE}_{\max } \\ & \mathrm{KE}_{\max }=\mathrm{eV}_0 \\ & \mathrm{~V}_0=\frac{\mathrm{hv}-\phi}{\mathrm{e}} \end{aligned}$

(A) $\mathrm{V}_0 \mathrm{v} / \mathrm{s} \mathrm{V}$ is linear correct

(B) Slope

$\mathrm{v}_0=\left(\frac{\mathrm{h}}{\mathrm{e}}\right) \mathrm{v}-\frac{\phi}{\mathrm{e}} \text { Wrong }$

Slope $\frac{\mathrm{h}}{\mathrm{e}}$

(C) Correct

(D) Incorrect

(E) Correct

Derivation of the de Broglie wavelength after time $t$

The de Broglie wavelength is given by

$ \lambda = \frac{h}{p}, $

where $h$ is Planck's constant and $p$ is the momentum of the particle.

Initial Conditions:

Initially, the electron has a velocity

$ \vec{v} = v_0\,\hat{i}, $

so its momentum is

$ p_0 = m v_0. $

The initial de Broglie wavelength is therefore

$ \lambda_0 = \frac{h}{m v_0}. $

Effect of the Electric Field:

The electron enters an electric field

$ \vec{E} = -E_0\,\hat{k}. $

For an electron with charge $-e$, the force is given by

$ \vec{F} = -e\,\vec{E} = -e\left(-E_0\,\hat{k}\right) = e E_0\,\hat{k}. $

This force causes an acceleration in the $\hat{k}$ direction:

$ a_z = \frac{F_z}{m} = \frac{e E_0}{m}. $

Momentum Components After Time $t$:

In the $\hat{i}$ direction, there is no force, so the momentum remains constant:

$ p_x = m v_0. $

In the $\hat{k}$ direction, starting from rest, the momentum becomes:

$ p_z = m v_z = m \left(a_z t\right) = e E_0 t. $

Total Momentum of the Electron:

The total momentum is the vector sum of the components:

$ p = \sqrt{p_x^2 + p_z^2} = \sqrt{(m v_0)^2 + (e E_0 t)^2}. $

New de Broglie Wavelength:

Substituting the expression for the total momentum into the de Broglie relation gives:

$ \lambda = \frac{h}{\sqrt{m^2 v_0^2 + e^2 E_0^2 t^2}}. $

Expressing this in terms of the initial de Broglie wavelength $\lambda_0 = \frac{h}{m v_0}$:

$ \lambda = \frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}. $

This final expression matches the option:

$ \frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}. $

Let's break down the problem step by step:

In the photoelectric effect, the energy of a photon is given by:

$ h\nu = \phi + K_e $

where:

$\phi$ is the work function of the metal,

$K_e$ is the kinetic energy of the ejected electron.

The kinetic energy of the ejected electrons is provided by the stopping potential:

$ K_e = eV $

Given that the stopping potential is $ V = 2 \text{ V} $, we have:

$ K_e = 2 \text{ eV} $

Now, we can calculate the photon energy:

$ h\nu = \phi + K_e = 2.14 \text{ eV} + 2 \text{ eV} = 4.14 \text{ eV} $

The relation between the photon's energy and its wavelength is:

$ h\nu = \frac{hc}{\lambda} $

Rearranging for $\lambda$:

$ \lambda = \frac{hc}{h\nu} $

Given that $ hc = 1242 \text{ eV·nm} $, substitute the values:

$ \lambda = \frac{1242 \text{ eV·nm}}{4.14 \text{ eV}} $

Calculate the wavelength:

$ \lambda \approx \frac{1242}{4.14} \approx 300 \text{ nm} $

Thus, the wavelength of the incident electromagnetic wave is approximately 300 nm.

The correct answer is Option C: 300 nm.

To find the matter wave behavior of the particle, we use the de Broglie wavelength formula:

$\lambda = \frac{h}{mv}$

where:

$h = 6.63 \times 10^{-34} \, \mathrm{J \cdot s}$ is Planck’s constant,

$m = 10^{-30} \, \mathrm{kg}$ is the mass of the particle, and

$v = 2.21 \times 10^6 \, \mathrm{m/s}$ is its velocity.

Follow these steps:

Substitute the values into the formula:

$\lambda = \frac{6.63 \times 10^{-34}}{(10^{-30})(2.21 \times 10^6)}$

Calculate the denominator:

$10^{-30} \times 2.21 \times 10^6 = 2.21 \times 10^{-24}$

Now compute the wavelength:

$\lambda \approx \frac{6.63 \times 10^{-34}}{2.21 \times 10^{-24}} \approx 3 \times 10^{-10} \, \mathrm{m}$

This wavelength, roughly $3 \times 10^{-10} \, \mathrm{m}$ (or 0.3 nm), lies within the X-ray portion of the electromagnetic spectrum, since X-rays typically have wavelengths in the range of about 0.01 nm to 10 nm.

Therefore, under the matter wave consideration, the particle will behave closely like:

Option A: X-rays.

Let's break down the problem step by step.

The photoelectric equation is given by:

$K_{\text{max}} = \frac{hc}{\lambda} - \phi$

where:

$K_{\text{max}}$ is the maximum kinetic energy of the electrons.

$\frac{hc}{\lambda}$ is the energy of the incident photon.

$\phi$ is the work function of the metal.

For the initial light source of wavelength $\lambda$, we know:

$2 \text{ eV} = \frac{hc}{\lambda} - 1 \text{ eV}$

Solving for $\frac{hc}{\lambda}$ gives:

$\frac{hc}{\lambda} = 2 \text{ eV} + 1 \text{ eV} = 3 \text{ eV}$

Now, if the wavelength is halved to $\frac{\lambda}{2}$, the photon energy becomes:

$\frac{hc}{\lambda/2} = \frac{2hc}{\lambda} = 2 \times 3 \text{ eV} = 6 \text{ eV}$

The maximum kinetic energy for the new wavelength is then:

$K'_{\text{max}} = 6 \text{ eV} - 1 \text{ eV} = 5 \text{ eV}$

Therefore, the maximum kinetic energy of the ejected electrons when the surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$ is 5 eV.

The correct answer is Option A.

For photo-electric effect, energy of the photon must be greater than the work function of the metal.

We know, $E = {{1240} \over \lambda } = {{1240} \over {550}}$ (as $\lambda=550$ nm given)

$\Rightarrow E=2.25$ eV

here, $E > {\phi _{cs}}$

So, $C_S$ only.

We know, $\lambda = {h \over {mv}}$

and $mvr = {{nh} \over {2\pi }}$

$ \Rightarrow mv = {{nh} \over {2\pi r}}$

So, $\lambda = {h \over {nh}}2\pi r = 2\pi {r \over n}$

$ \Rightarrow \lambda \propto {r \over n}$

We can write, ${{{\lambda _g}} \over {{\lambda _e}}} = \left( {{{{r_g}} \over {{r_e}}}} \right)\left( {{{{n_e}} \over {{n_g}}}} \right)$

$ \Rightarrow {{{\lambda _g}} \over {{\lambda _e}}} = \left( {{{5.3 \times {{10}^{ - 11}}} \over {84.8 \times {{10}^{ - 11}}}}} \right)\left( {{4 \over 1}} \right) = {1 \over 4}$

$ \Rightarrow {{{\lambda _e}} \over {{\lambda _g}}} = 4$

To find the maximum kinetic energy of the ejected photoelectrons, we'll use the photoelectric effect equation:

where $KE_{\text{max}}$ is the maximum kinetic energy of the ejected electrons, $h$ is Planck's constant, $\nu$ is the frequency of the incident light, and $\phi$ is the work function of the metal.

However, in this problem, we are given the energy of the UV light in electronvolts (eV) directly, which simplifies the problem. The energy of the UV light in electronvolts also represents the energy of the photons ($h\nu$) hitting the metal surface. Thus, we can calculate the maximum kinetic energy of the ejected photoelectrons using the given energies directly:

Substituting the given values:

Therefore, the maximum kinetic energy of the ejected photoelectrons will be $1 \, \text{eV}$. The correct answer is Option B: 1 eV.

To determine the relationship between the de-Broglie wavelengths of a proton, an electron, and an alpha particle with the same energies, we need to use the de-Broglie wavelength formula:

$\lambda = \frac{h}{p}$

where $ h $ is Planck's constant and $ p $ is the momentum of the particle.

For particles with the same kinetic energy $ E $, we have:

$E = \frac{p^2}{2m}$

Solving for $ p $:

$p = \sqrt{2mE}$

Substituting this into the de-Broglie equation, we get:

$\lambda = \frac{h}{\sqrt{2mE}}$

Since all three particles have the same energy $ E $, the de-Broglie wavelength is inversely proportional to the square root of the mass $ m $:

$\lambda \propto \frac{1}{\sqrt{m}}$

The masses of the particles are as follows:

• Mass of proton $ m_p $ is approximately $ 1.67 \times 10^{-27} $ kg
• Mass of electron $ m_e $ is approximately $ 9.11 \times 10^{-31} $ kg
• Mass of alpha particle $ m_{\alpha} $ is approximately $ 4 \times 1.67 \times 10^{-27} $ kg (since it has 2 protons and 2 neutrons)

Comparatively:

• Mass of alpha particle $ m_{\alpha} $ is the largest.
• Mass of proton $ m_p $ is intermediate.
• Mass of electron $ m_e $ is the smallest.

Therefore, the de-Broglie wavelength will be:

• Largest for the electron ($ \lambda_{\mathrm{e}} $)
• Intermediate for the proton ($ \lambda_{\mathrm{p}} $)
• Smallest for the alpha particle ($ \lambda_{\alpha} $)

Hence, the correct order is:

Option B:

$\lambda_\alpha<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}$

To determine the correct relation between the kinetic energies of a proton ($\mathrm{K}_{\mathrm{p}}$) and an electron ($\mathrm{K}_{\mathrm{e}}$) when they have the same de Broglie wavelength, we need to use the de Broglie wavelength formula:

$\lambda = \frac{h}{p}$

where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, and $p$ is the momentum of the particle.

The momentum $p$ of a particle is given by:

$p = \sqrt{2mK}$

where $m$ is the mass of the particle and $K$ is its kinetic energy. For a proton and an electron with the same de Broglie wavelength:

$\lambda_{\mathrm{p}} = \lambda_{\mathrm{e}}$

This implies the momenta should be the same:

$p_{\mathrm{p}} = p_{\mathrm{e}}$

Thus, we can write:

$\sqrt{2 m_{\mathrm{p}} K_{\mathrm{p}}} = \sqrt{2 m_{\mathrm{e}} K_{\mathrm{e}}}$

Squaring both sides to eliminate the square roots:

$2 m_{\mathrm{p}} K_{\mathrm{p}} = 2 m_{\mathrm{e}} K_{\mathrm{e}}$

$m_{\mathrm{p}} K_{\mathrm{p}} = m_{\mathrm{e}} K_{\mathrm{e}}$

Rearranging to solve for $K_{\mathrm{p}}$ in terms of $K_{\mathrm{e}}$:

$K_{\mathrm{p}} = \frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} K_{\mathrm{e}}$

Since the mass of a proton $m_{\mathrm{p}}$ is much greater than the mass of an electron $m_{\mathrm{e}}$:

$m_{\mathrm{p}} \gg m_{\mathrm{e}}$

This means:

$\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} \ll 1$

Therefore, it implies:

$K_{\mathrm{p}} < K_{\mathrm{e}}$

So, the correct option is:

Option C: $\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}$

To solve for the ratio of the kinetic energies of a proton and an electron with the same de-Broglie wavelength, let us first recall the relationship between kinetic energy, momentum, and the de-Broglie wavelength.

The de-Broglie wavelength $\lambda$ is given by:

$\lambda = \frac{h}{p}$

where $h$ is Planck’s constant and $p$ is the momentum.

Since the proton and the electron have the same de-Broglie wavelength, their momenta must be equal:

$\lambda_{\text{electron}} = \lambda_{\text{proton}} \implies \frac{h}{p_{\text{e}}} = \frac{h}{p_{\text{p}}} \implies p_{\text{e}} = p_{\text{p}}$

Next, the kinetic energy $K$ of a particle is related to its momentum $p$ and mass $m$ by the following formula:

$K = \frac{p^2}{2m}$

Given that the momentum $p$ is the same for both the proton and the electron, we can write the kinetic energies as:

$K_{\text{e}} = \frac{p^2}{2m_{\text{e}}}$

$K_{\text{p}} = \frac{p^2}{2m_{\text{p}}}$

The ratio of the kinetic energies is therefore:

$\frac{K_{\text{e}}}{K_{\text{p}}} = \frac{\frac{p^2}{2m_{\text{e}}}}{\frac{p^2}{2m_{\text{p}}}} = \frac{m_{\text{p}}}{m_{\text{e}}}$

Given that the mass of the proton $m_{\text{p}}$ is 1836 times the mass of the electron $m_{\text{e}}$, we have:

$\frac{m_{\text{p}}}{m_{\text{e}}} = 1836$

Thus, the ratio of the kinetic energies is:

$\frac{K_{\text{e}}}{K_{\text{p}}} = 1836$

Therefore, the correct answer is:

Option C

$1: 1836$

To find the stopping potential ($V_s$) when UV light of wavelength $300 $ nm is incident on a metal surface with a work function of $2.13$ eV, we can use the photoelectric equation which relates the energy of the incident photons, the work function of the metal, and the kinetic energy of the emitted electrons.

The energy (E) of the photons can be calculated using the equation:

$E = \frac{hc}{\lambda}$

where $h$ is the Planck constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the incident light. Given that $hc = 1240 $ eV nm, we can calculate the energy of the UV light photons directly.

Substituting $hc = 1240 $ eV nm and $\lambda = 300 $ nm into the equation gives:

$E = \frac{1240 \, \text{eV nm}}{300 \, \text{nm}} = 4.13 \, \text{eV}$

Next, we can calculate the maximum kinetic energy of the emitted electrons using the photoelectric effect equation:

$K_{max} = E - \phi$

where $K_{max}$ is the maximum kinetic energy of the emitted electrons, $E$ is the energy of the incident photons, and $\phi$ is the work function of the metal.

Given $E = 4.13$ eV and the work function $\phi = 2.13$ eV, we have:

$K_{max} = 4.13 \, \text{eV} - 2.13 \, \text{eV} = 2 \, \text{eV}$

The stopping potential ($V_s$) is related to the maximum kinetic energy of the emitted electrons by the equation:

$K_{max} = eV_s$

where $e$ is the elementary charge (the charge of an electron), and $V_s$ is the stopping potential. Since $e = 1$ when using energy in eV and potential in volts, the stopping potential $V_s$ can be directly equated to the kinetic energy in eV:

$V_s = K_{max} = 2 \, \text{V}$

Therefore, the stopping potential is $2$ V, which corresponds to Option B.

The correct answer is Option A: C only.

Reflection, diffraction, interference, and polarization are all phenomena that can be explained by the wave nature of light. These phenomena are evidence that light behaves as a wave, evident through various experimental observations:

• Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. The laws of reflection can be explained by treating light as waves.
• Diffraction is the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle. This phenomenon is observed when a wave encounters an obstacle or a slit that is comparable in size to its wavelength, which is best explained by wave theory.
• Interference occurs when two or more waves superimpose to form a resultant wave of greater, lower, or the same amplitude. This phenomenon illustrates the wave nature of light through constructive and destructive interference patterns, such as those observed in the double-slit experiment.
• Polarization is a property applying to transverse waves that specifies the geometrical orientation of the oscillations. For light waves, this can be seen as the direction in which the electric field oscillates. Polarization can only occur with waves that have transverse components, further supporting the wave nature of light.

On the other hand, the photoelectric effect cannot be explained solely by the wave nature of light. It is the emission of electrons or other free carriers when light shines on a material. Electrons emitted in this manner can be called photoelectrons. The phenomenon is best explained by Albert Einstein's quantum theory of light, where light is considered as quanta of energy called photons. This effect demonstrates the particle aspect of light, wherein each photon has a discrete packet of energy equal to $hf$, where $h$ is Planck's constant and $f$ is the frequency of the light.

Thus, the photoelectric effect is the correct answer because it specifically requires the particle theory of light for its explanation, unlike reflection, diffraction, interference, and polarization, which are well-explained by the wave nature of light.

To find the work function of the photo sensitive material, we can use the photoelectric equation which relates the kinetic energy of the ejected electrons to the photon energy and the work function ($\phi$) of the material:

$KE_{\text{max}} = h\nu - \phi$

Where $KE_{\text{max}}$ is the maximum kinetic energy of the ejected electrons, $h\nu$ is the energy of the incoming photon, and $\phi$ is the work function of the material.

However, the kinetic energy of the ejected electrons can also be related to the stopping potential ($V_s$) by the equation:

$KE_{\text{max}} = e \cdot V_s$

Substituting this into the first equation gives:

$e \cdot V_s = h\nu - \phi$

Here, $e$ is the charge of an electron ($1.6 \times 10^{-19} \, \text{C}$), but since we are dealing with energies in electronvolts (eV), and $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$, we can directly use the values given without converting the units:

$\phi = h\nu - e \cdot V_s$

Where $h\nu = 2.48 \, \text{eV}$ is the energy of the irradiating photons, and $V_s = 0.5 \, \text{V}$.

Substituting these values in, we get:

$\phi = 2.48 \, \text{eV} - 0.5 \, \text{eV} = 1.98 \, \text{eV}$

Therefore, the work function ($\phi$) of the photo sensitive material is $1.98 \, \text{eV}$, which corresponds to Option A.

Stopping potential is independent of intensity of light. It depends on frequency of light.

$\begin{aligned} & V_0=\frac{h}{e} f-\frac{h}{e} f_0 \\ & \Rightarrow \text { Slope }=\frac{h}{e} \text { (S-I is correct) } \\ & \because\left(f_0\right)_1<\left(f_0\right)_2 \\ & \therefore \quad(\mathrm{S}-I I \text { is incorrect) } \\ & \therefore \quad \text { Option (3) is correct. } \end{aligned}$

In order to determine the most appropriate answer to the question, let's analyze the given Assertion A and Reason R in detail.

Assertion A: Number of photons increases with increase in frequency of light.

This statement is not correct. The number of photons is determined by the intensity (or power) of the light and is given by the formula:

$N = \frac{P}{hf}$

where $N$ is the number of photons per second, $P$ is the power (intensity) of the light, $h$ is Planck's constant, and $f$ is the frequency of the light. As the frequency increases, the energy per photon increases, but it does not necessarily mean that the number of photons increases unless the power of the light also increases proportionally.

Reason R: Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.

This statement is correct. According to the photoelectric effect, the maximum kinetic energy of emitted electrons is given by:

$K.E_{\text{max}} = hf - \phi$

where $K.E_{\text{max}}$ is the maximum kinetic energy of the emitted electrons, $h$ is Planck's constant, $f$ is the frequency of the incident radiation, and $\phi$ is the work function of the material. As the frequency of the incident light increases, the kinetic energy of the emitted electrons increases.

Now, let's match the statements with the options given:

Option A: $\mathbf{A}$ is not correct but $\mathbf{R}$ is correct.

This option is correct because Assertion A is incorrect while Reason R is correct.

Option B: $\mathbf{A}$ is correct but $\mathbf{R}$ is not correct.

This option is incorrect because Assertion A is not correct.

Option C: Both $\mathbf{A}$ and $\mathbf{R}$ are correct and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$.

This option is incorrect because Assertion A is not correct.

Option D: Both $\mathbf{A}$ and $\mathbf{R}$ are correct and $\mathbf{R}$ is NOT the correct explanation of $\mathbf{A}$.

This option is incorrect because Assertion A is not correct.

Therefore, the most appropriate answer is:

Option A: $\mathbf{A}$ is not correct but $\mathbf{R}$ is correct.

Stopping potential is independent on intensity but photocurrent increases non-linearly on increasing intensity.

To find out the number of photons emitted per second by the laser, we can use the relationship between the energy of a single photon, the total energy emitted per second (power), and the number of photons emitted per second. The energy $E$ of a single photon is given by Planck's equation:

$ E = hf $

where:

• $ h $ is Planck's constant ($ 6.63 \times 10^{-34} \mathrm{Js} $), and
• $ f $ is the frequency of the light ($ 6 \times 10^{14} \mathrm{Hz} $).

Let's first calculate the energy of one photon:

$ E = (6.63 \times 10^{-34} \mathrm{Js}) \times (6 \times 10^{14} \mathrm{Hz}) $

$ E = 3.978 \times 10^{-19} \mathrm{J} $

The power ($ P $) emitted by the laser is the total energy emitted per second,

$ P = E_{\text{total per second}} = 2 \times 10^{-3} \mathrm{W} = 2 \times 10^{-3} \mathrm{J/s} $

The number of photons ($ N $) emitted per second can be found by dividing the total energy emitted per second by the energy of one photon:

$ N = \frac{P}{E} $

Substitute the values we have:

$ N = \frac{2 \times 10^{-3} \mathrm{J/s}}{3.978 \times 10^{-19} \mathrm{J}} $

$ N = \frac{2 \times 10^{-3}}{3.978 \times 10^{-19}} $

$ N = 5.03 \times 10^{15} \text{ photons per second} $

The number of photons emitted per second is approximately $5 \times 10^{15}$. Therefore, the correct answer, rounded to one significant figure, is:

Option A: $5 \times 10^{15}$

To find the value of $ X $ in the band gap $ \left(\frac{\mathrm{X}}{8}\right) \mathrm{eV} $, we need to understand the relationship between the wavelength of light that can result in changes in the conductivity of a photodiode and the photodiode's band gap energy.

The band gap energy ($ E_{g} $) of a material is the minimum energy required for an electron to transition from the valence band to the conduction band, thus creating a hole-electron pair and allowing conductivity to occur. When light with a certain wavelength ($ \lambda $) is incident upon the photodiode, if the energy of the photons is greater than or equal to the band gap energy of the photodiode, then the photons can excite electrons and change the conductivity of the photodiode.

The energy of a photon ($ E_{photon} $) is given by the equation:

$ E_{photon} = \frac{hc}{\lambda} $

where:

• $ h $ is Planck's constant, which is given as $ 6.6 \times 10^{-34} $ J·s.
• $ c $ is the speed of light in a vacuum ($ \approx 3 \times 10^{8} $ m/s).
• $ \lambda $ is the wavelength of the incident light.

Then the band gap energy in terms of electron volts is found by converting the energy from joules to electron volts (eV) using the charge of an electron $ e $ ($ 1.6 \times 10^{-19} $ C):

$ E_{g} = \frac{hc}{\lambda e} $

Given that the photodiode starts conducting when the wavelength of light is less than $ 660 \mathrm{~nm} $, we should use this wavelength as the threshold wavelength $ \lambda_{threshold} $:

$ E_{g} = \frac{hc}{\lambda_{threshold}\times e} $

$ E_{g} = \frac{6.6 \times 10^{-34} \text{ J·s} \cdot 3 \times 10^{8} \text{ m/s}}{660 \times 10^{-9} \text{ m} \times 1.6 \times 10^{-19} \text{ C}} $

$ E_{g} = \frac{6.6 \times 3 \times 10^{-26}}{660 \times 1.6} \mathrm{eV} $

$ E_{g} = \frac{19.8}{660 \times 1.6} \mathrm{eV} $

$ E_{g} = \frac{19.8}{1056} \mathrm{eV} $

$ E_{g} = \frac{18.75}{1000} \mathrm{eV} $

$ E_{g} = 1.875 \mathrm{eV} $

Now, we have to compare this value to $ \left(\frac{\mathrm{X}}{8}\right) \mathrm{eV} $ to find the value of $ X $:

$ 1.875 = \frac{X}{8} $

$ X = 1.875 \times 8 $

$ X = 15 $

Therefore, the value of $ X $ is 15. The correct answer is Option C.

To find the ratio of velocities of two particles based on their de Broglie wavelengths, we can use the de Broglie wavelength formula, which relates the momentum of a particle to its wavelength. The de Broglie's wavelength formula is given by:

$ \lambda = \frac{h}{p} $

where:

$\lambda$ is the de Broglie wavelength,

$h$ is the Planck constant, and

$p$ is the momentum of the particle.

The momentum $p$ of a particle can also be expressed as the product of its mass $m$ and velocity $v$:

$ p = mv $

So, we can rewrite the de Broglie wavelength equation in terms of mass and velocity:

$ \lambda = \frac{h}{mv} $

For the proton (let's use subscript $p$ for proton), the wavelength is $\lambda$:

$ \lambda_p = \frac{h}{m_p v_p}$

For the $\alpha$ particle (let's use subscript $\alpha$ for alpha), the wavelength is $2\lambda$:

$ 2\lambda = \frac{h}{m_\alpha v_\alpha}$

We are interested in finding the ratio of the velocities $\frac{v_p}{v_\alpha}$. Using the given data about wavelengths:

$ \lambda_p = \lambda $

$ \lambda_\alpha = 2\lambda $

Using the de Broglie equation for both particles:

$ \frac{h}{m_p v_p} = \lambda $

$ \frac{h}{m_\alpha v_\alpha} = 2\lambda $

Dividing the second equation by the first equation gives us:

$ \frac{\frac{h}{m_\alpha v_\alpha}}{\frac{h}{m_p v_p}} = \frac{2\lambda}{\lambda} $

$ \frac{m_p v_p}{m_\alpha v_\alpha} = \frac{2}{1} $

$ \frac{v_p}{v_\alpha} = \frac{2m_\alpha}{m_p} $

Since we know an $\alpha$ particle consists of 2 protons and 2 neutrons (essentially four nucleons), the mass of an $\alpha$ particle is roughly four times the mass of a proton ($m_\alpha \approx 4m_p$).

Substituting $m_\alpha$ with $4m_p$ in the equation:

$ \frac{v_p}{v_\alpha} = \frac{2(4m_p)}{m_p} $

$ \frac{v_p}{v_\alpha} = 2 \cdot 4 $

$ \frac{v_p}{v_\alpha} = 8 $

Therefore, the ratio of the velocities of proton to $\alpha$ particle is $8:1$, which corresponds to Option A.

Since $\frac{\mathrm{f}}{2}<\mathrm{f}_0$

i.e. the incident frequency is less than threshold frequency. Hence there will be no emission of photoelectrons.

$\Rightarrow \text { current }=0$

$\begin{aligned} & \mathrm{E}=\phi+\mathrm{K}_{\max } \\ & \phi=\frac{\mathrm{hc}}{\lambda_0} \\ & \mathrm{~K}_{\max }=\mathrm{eV}_0 \\ & 8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . \text { (i) } \\ & 2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . . \text { (ii) } \\ & \text { on solving (i) & (ii) } \\ & \lambda_0=9 \lambda \end{aligned}$

$\begin{aligned} & \text { K.E. }=\mathrm{hf}-\phi \\ & \tan \theta=\mathrm{h} \end{aligned}$

$\mathrm{p=\frac{E}{C}=\frac{6.48 \times 10^5}{3 \times 10^8}=2.16 \times 10^{-3}}$

$\begin{aligned} & \text { For P.E.E. : } \lambda \leq \frac{h c}{W_e} \\ & \lambda \leq \frac{1240 \mathrm{~nm}-\mathrm{eV}}{3 \mathrm{eV}} \\ & \lambda \leq 413.33 \mathrm{~nm} \\ & \lambda_{\max } \approx 414 \mathrm{~nm} \text { for P.E.E. } \end{aligned}$

$\begin{aligned} & \mathrm{n}_1 \times \frac{\mathrm{hc}}{\lambda_1}=200 \\ & \mathrm{n}_2 \times \frac{\mathrm{hc}}{\lambda_2}=200 \\ & \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{\lambda_1}{\lambda_2}=\frac{300}{500} \\ & \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{3}{5} \end{aligned}$

We know that the de-Broglie wavelength $\lambda$ of a particle is given by:

$\lambda = \frac{h}{p}$

where:

• $h$ is Planck's constant,
• $p$ is the momentum of the particle.

For a photon (which has zero rest mass), its energy $E$ and momentum $p$ are related by the equation:

$E = cp$

and its de-Broglie wavelength $\lambda$ is given by:

$\lambda = \frac{h}{E} \times \frac{1}{c}$

Now, since the photon and electron are said to have the same de-Broglie wavelength:

$\lambda_{electron} = \lambda_{photon}$

$\frac{h}{p_{electron}} = \frac{h}{E_{photon}} \times \frac{1}{c}$

For the electron, its momentum $p_{electron}$ is given by:

$p_{electron} = m_{e}v_{electron}$

where:

• $m_{e}$ is the electron's rest mass,
• $v_{electron}$ is the electron's velocity.

The kinetic energy $K.E.$ of the electron is:

$K.E._{electron} = \frac{1}{2}m_{e}v_{electron}^2$

The question states that $v_{electron}$ is $25\%$ ($0.25c$) of the speed of light $c$. So we write:

$v_{electron} = 0.25c$

Plugging this into the kinetic energy formula, we get:

$K.E._{electron} = \frac{1}{2}m_{e}(0.25c)^2 = \frac{1}{2}m_{e} \times \frac{1}{16}c^2$

$K.E._{electron} = \frac{1}{32}m_{e}c^2$

For a photon, $p_{photon} = \frac{E_{photon}}{c}$ and hence its kinetic energy (which, for a photon, is simply its energy) is:

$K.E._{photon} = E_{photon} = cp_{photon}$

Now, comparing the kinetic energies:

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}m_{e}c^2}{cp_{photon}}$

Since $p_{electron} = p_{photon}$ (from the de-Broglie relation), we can replace $p_{photon}$ with $p_{electron}$ which is $m_{e}v_{electron}$:

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}m_{e}c^2}{m_{e}v_{electron}c}$

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}c}{0.25c}$

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{32} \times \frac{1}{0.25}$

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{32} \times 4$

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{8}$

So the correct answer is Option C $\frac{1}{8}$.

The threshold frequency, $ \nu_0 $, corresponds to the minimum frequency of light required to eject electrons from the surface of a metal, a phenomenon known as the photoelectric effect. The work function, represented by $ \phi $, is the minimum energy needed to remove an electron from the surface of the metal.

The energy of a photon is given by the equation $ E = h\nu $, where:

• $ E $ is the energy of the photon,
• $ h $ is Planck's constant ($ h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} $), and
• $ \nu $ is the frequency of the photon.

To find the threshold frequency for a metal with a work function of $ 6.63 \, \text{eV} $, we must first express the work function in joules (since Planck's constant is in joules per second). To convert electron volts to joules, use the conversion factor $ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} $:

The energy of the photon at the threshold frequency is equal to the work function:

Solve for $ \nu_0 $:

Thus, the correct answer is:

Threshold wavelength ($\lambda_{threshold}$) is the maximum wavelength of light required to remove an electron from a metal surface, i.e., to overcome the work function ($\phi$). The relationship between work function and threshold wavelength is given by:

Where:

• $\phi$ is the work function in electron-volts (eV)
• $h$ is the Planck's constant
• $c$ is the speed of light
• $\lambda_{threshold}$ is the threshold wavelength

In this problem, we are given the product $hc = 1242 ~\mathrm{eV}\,\mathrm{nm}$, and the work functions $\phi_{A} = 9 ~\mathrm{eV}$ and $\phi_{B} = 4.5 ~\mathrm{eV}$. Let's calculate the threshold wavelengths for metal surfaces $\mathrm{A}$ and $\mathrm{B}$:

For metal surface $\mathrm{A}$:

For metal surface $\mathrm{B}$:

Now let's calculate the difference between the threshold wavelengths ($\Delta\lambda$):

By calculating the difference, we get:

So, the difference between the threshold wavelengths for the two metal surfaces is $\boxed{138}\,\mathrm{nm}$.