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).

The power $(\mathrm{P})$ of a light source is the total energy emitted per unit time. If the source emits n photons per second, and each photon has an energy E , then the power is:

$ \mathrm{P}=\mathrm{n} \times \mathrm{E} $

The given power of the source is $\mathrm{P}=15 \mathrm{~kW}=15 \times 10^3 \mathrm{~W}$ and the number of photons per second is $\mathrm{n}= 2.5 \times 10^{22} \mathrm{~s}^{-1}$

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

$\Rightarrow $ $\mathrm{E}=\frac{15 \times 10^3}{2.5 \times 10^{22}}$

$\Rightarrow $ $ \mathrm{E}=6 \times 10^{-19} \mathrm{~J} $

The energy of a photon is related to its wavelength by the formula:

$ E=\frac{h c}{\lambda} \Rightarrow \lambda=\frac{h c}{E} $

Substituting the constants and energy,

The Plank's constant is $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s}$,

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

$ \lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6 \times 10^{-19}} \mathrm{~m} $

$\Rightarrow $ $\lambda=\frac{19.8 \times 10^{-26}}{6 \times 10^{-19}} \mathrm{~m}$

$\Rightarrow $ $\lambda=3.3 \times 10^{-7} \mathrm{~m}$

$\Rightarrow $ $ \lambda=3.3 \times 10^{-7} \times 10^9 \mathrm{~nm}=330 \mathrm{~nm} $

Visible light typically ranges from approximately 400 nm to 700 nm . And the ultraviolet (UV) radiation ranges from approximately 10 nm to 400 nm . Since 330 nm is less than 400 nm , the radiation belongs to the ultraviolet region.

Therefore, the emitted radiation belongs to the Ultraviolet region.

Hence, the correct option is (D).

According to the photoelectric effect, when light of a certain frequency or wavelength hits a metal surface, it provides energy to the electrons. Part of this energy is used to overcome the attractive forces of the metal (the work function, $\phi$ ), and the remaining energy becomes the maximum kinetic energy (K) of the emitted photoelectron. The energy of an incident photon is $E=\frac{h c}{\lambda}$. The equation is:

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

$\Rightarrow $ $\frac{\mathrm{hc}}{\lambda}=\phi+\mathrm{K}_{\max }$

$\Rightarrow $ $ \mathrm{K}_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi $

We are given two different wavelengths and their corresponding kinetic energies:

For wavelength $\lambda_1$ :

$ \begin{equation*} \mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_1}-\phi \ldots \tag{i} \end{equation*} $

For wavelength $\lambda_2$ :

$ \begin{equation*} \mathrm{K}_2=\frac{\mathrm{hc}}{\lambda_2}-\phi \ldots \tag{ii} \end{equation*} $

It is given that $\lambda_1=2 \lambda_2$.

$ \begin{equation*} \mathrm{K}_1=\frac{\mathrm{hc}}{2 \lambda_2}-\phi \ldots \tag{iii} \end{equation*} $

From equation (ii), we can find an expression for $\frac{\mathrm{hc}}{\lambda_2}$ :

$ \frac{\mathrm{hc}}{\lambda_2}=\mathrm{K}_2+\phi $

Now, substituting this into equation (iii) :

$ \mathrm{K}_1=\frac{1}{2}\left(\mathrm{~K}_2+\phi\right)-\phi $

$\Rightarrow $ $2 \mathrm{~K}_1=\mathrm{K}_2+\phi-2 \phi$

$\Rightarrow $ $2 \mathrm{~K}_1=\mathrm{K}_2-\phi$

$\Rightarrow $ $ \phi=\mathrm{K}_2-2 \mathrm{~K}_1 $

Therefore, the work function of the material is $K_2-2 K_1$.

Hence, the correct option is (D).

For an electron, the de Broglie wavelength is given by

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

where $h$ is Planck’s constant, $m$ is the mass of the electron, and $v$ is its velocity.

In free space, its wavelength is

$ \lambda_0 = \frac{h}{mv} $

When the electron enters the medium, its velocity is reduced by $20\%$.

So the new velocity becomes

$ v' = v - 0.2v = 0.8v $

Now the new de Broglie wavelength in the medium is

$ \lambda = \frac{h}{m v'} = \frac{h}{m(0.8v)} $

$ \lambda = \frac{1}{0.8}\cdot \frac{h}{mv} $

$ \lambda = 1.25 \lambda_0 $

Comparing with

$ \lambda = \alpha \lambda_0 $

we get

$ \alpha = 1.25 $

So, the correct answer is:

$ \boxed{1.25} $

Option C

For an electron, charge is $q=-e$.

Given electric field,

$ \vec E=-2E_0 \hat i $

Force on the electron is

$ \vec F=q\vec E=(-e)(-2E_0 \hat i)=2eE_0 \hat i $

So acceleration is

$ \vec a=\frac{\vec F}{m}=\frac{2eE_0}{m}\hat i $

Since initial velocity is along $\hat i$,

$ \vec v(0)=v_0 \hat i $

therefore velocity at time $t$ will be

$ \vec v(t)=\left(v_0+\frac{2eE_0}{m}t\right)\hat i $

Now de Broglie wavelength is

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

So at time $t$,

$ \lambda(t)=\frac{h}{m\left(v_0+\frac{2eE_0}{m}t\right)} $

Given

$ \lambda_0=\frac{h}{4mv_0} $

So,

$ \frac{h}{mv_0}=4\lambda_0 $

Hence,

$ \lambda(t)=\frac{\frac{h}{mv_0}}{1+\frac{2eE_0}{m}\frac{t}{v_0}} =\frac{4\lambda_0}{1+\frac{2eE_0}{m}\frac{t}{v_0}} $

Therefore the correct answer is

$ \boxed{\frac{4\lambda_0}{\left[1+\frac{2E_0e}{m}\frac{t}{v_0}\right]}} $

So, Option C is correct.

Use Einstein’s photoelectric equation:

$ h\nu=\phi+K_{\max} $

Here,

• $\phi$ = work function

• $K_{\max}=eV_0$

• $V_0=0.2\ \text{V}$

Also,

$ \nu=\frac{c}{\lambda} $

So,

$ \phi=\frac{hc}{\lambda}-eV_0 $

Now substitute the values:

$ h=6.62\times 10^{-34}\ \text{J s},\quad c=3\times 10^8\ \text{m/s},\quad \lambda=331\ \text{nm}=331\times 10^{-9}\ \text{m} $

First, calculate photon energy:

$ \frac{hc}{\lambda} = \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}} = \frac{19.86}{331}\times 10^{-17} $

$ \approx 0.06\times 10^{-17} =6\times 10^{-19}\ \text{J} $

Now,

$ eV_0=(1.6\times 10^{-19})(0.2)=0.32\times 10^{-19}\ \text{J} $

Therefore,

$ \phi=6\times 10^{-19}-0.32\times 10^{-19} $

$ \phi=5.68\times 10^{-19}\ \text{J} $

Comparing with $\alpha\times 10^{-19}\ \text{J}$,

$ \alpha=5.68 $

Therefore, the correct answer is:

$ \boxed{5.68} $

Option C.

For a charged particle accelerated through a potential difference $V$, the kinetic energy gained is

$ K = qV $

For both electron and proton, the magnitude of charge is the same, that is $e$. So,

$ K = eV $

Now, using the non-relativistic expression for kinetic energy,

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

So,

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

This gives

$ p = \sqrt{2meV} $

From de Broglie relation,

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

Substituting $p$,

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

So, for the electron,

$ \lambda_e = \frac{h}{\sqrt{2m_e eV}} $

and for the proton,

$ \lambda_p = \frac{h}{\sqrt{2m_p eV}} $

Now taking the ratio,

$ \frac{\lambda_e}{\lambda_p} = \frac{h/\sqrt{2m_e eV}}{h/\sqrt{2m_p eV}} = \sqrt{\frac{m_p}{m_e}} $

Hence,

$ \boxed{\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}} $

So, the correct option is:

$ \boxed{\text{Option A}} $

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$

$ e V_s=h v=\phi_0 $

$ \begin{aligned} \Rightarrow \quad V_s & =\frac{h}{e} v-\frac{\phi_0}{e} \\ \therefore \text { Slope } & =\frac{h}{e}=\frac{6.6 \times 10^{-34}}{1.6 \times 10^{-19}} \\ & =4.125 \times 10^{-15} \mathrm{Js} \mathrm{C}^{-1} \end{aligned} $

Work Done = Change in Kinetic Energy

The work needed is the same as how much the electron's kinetic energy increases.

At the start, the electron is at rest, so its initial kinetic energy $\mathrm{KE}_i = 0$.

After speeding up, the electron’s final kinetic energy $\mathrm{KE}_f = W$, where $W$ is the work done.

So, $W = \mathrm{KE}_f - \mathrm{KE}_i = \mathrm{KE}_f = W$

Relating de-Broglie Wavelength to Kinetic Energy

The de-Broglie wavelength formula is:

$ \lambda_d = \frac{h}{mv} = \frac{h}{p} $

For an electron (or any particle), $p = \sqrt{2m\mathrm{KE}}$.

So, $ \lambda_d = \frac{h}{\sqrt{2m \mathrm{KE}}} $

From above, $\mathrm{KE} = W$ (work done).

Therefore, $ \lambda_d = \frac{h}{\sqrt{2mW}} $

Now, solve for $W$: $ \lambda_d^2 = \frac{h^2}{2mW} $ $ W = \frac{h^2}{\lambda_d^2 \cdot 2m} $

Plugging in the Numbers

Given:

• Planck's constant, $h = 6.6 \times 10^{-34}\ \mathrm{Js}$
• Mass of electron, $m = 9 \times 10^{-31}\ \mathrm{kg}$
• de-Broglie wavelength, $\lambda_d = 6600\ \text{Å} = 6.6 \times 10^{-7}\ \mathrm{m}$

So, $ W = \frac{(6.6 \times 10^{-34})^2}{(6.6 \times 10^{-7})^2 \times 2 \times 9.0 \times 10^{-31}}\ \mathrm{J} $

Now calculate each step:

Top: $(6.6 \times 10^{-34})^2 = 43.56 \times 10^{-68}$

Bottom: $(6.6 \times 10^{-7})^2 \times 2 \times 9.0 \times 10^{-31} = (43.56 \times 10^{-14}) \times 18.0 \times 10^{-31} = 784.08 \times 10^{-45}$

So, $ W = \frac{43.56 \times 10^{-68}}{784.08 \times 10^{-45}} $

This simplifies to: $ W = \frac{1}{18.0} \times 10^{-23}\ \mathrm{J} $

So, the answer is: $ W = 5.56 \times 10^{-25}\ \mathrm{J} $

The photoelectric effect equation relates the energy of incident photons, the work function of the material, and the maximum kinetic energy of emitted photoelectrons:

$E_{photon} = \phi + KE_{max}$

The maximum kinetic energy ($KE_{max}$) of the emitted photoelectrons is related to their maximum velocity ($v_{max}$) by the formula:

$KE_{max} = \frac{1}{2} m v_{max}^2$

The stopping potential ($V_s$) is the potential difference required to stop the most energetic photoelectrons. The work done by the stopping potential must be equal to the maximum kinetic energy of the electrons:

$KE_{max} = e V_s$

We are given:

• Work function ($\phi$) = 1.5 eV (Note: This is not directly needed to find the stopping potential if $v_{max}$ is given, but it would be needed to find the photon energy.)

• Maximum velocity ($v_{max}$) = $8 \times 10^5 \mathrm{~ms}^{-1}$

• Mass of electron ($m$) = $9 \times 10^{-31} \mathrm{~kg}$

• Charge of electron ($e$) = $1.6 \times 10^{-19} \mathrm{~C}$

First, calculate the maximum kinetic energy ($KE_{max}$) using the given maximum velocity:

$KE_{max} = \frac{1}{2} m v_{max}^2$

$KE_{max} = \frac{1}{2} \times (9 \times 10^{-31} \mathrm{~kg}) \times (8 \times 10^5 \mathrm{~ms}^{-1})^2$

$KE_{max} = \frac{1}{2} \times 9 \times 10^{-31} \times (64 \times 10^{10})$

$KE_{max} = \frac{1}{2} \times 9 \times 64 \times 10^{-31+10}$

$KE_{max} = 9 \times 32 \times 10^{-21}$

$KE_{max} = 288 \times 10^{-21} \mathrm{~J}$

Now, use the relationship between maximum kinetic energy and stopping potential ($KE_{max} = e V_s$) to find the stopping potential ($V_s$):

$V_s = \frac{KE_{max}}{e}$

$V_s = \frac{288 \times 10^{-21} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~C}}$

$V_s = \frac{288}{1.6} \times \frac{10^{-21}}{10^{-19}}$

$V_s = \frac{2880}{16} \times 10^{-2}$

$V_s = 180 \times 10^{-2}$

$V_s = 1.8 \mathrm{~V}$

The stopping potential of the photoelectrons is 1.8 V.

The final answer is $\boxed{\text{1.8 V}}$.

To find the smallest (minimum) wavelength of X-rays made by 20 kV electrons, use this formula:

$ \lambda_{min} = \frac{hc}{E} = \frac{hc}{eV} $ where:
$ h $ = Planck's constant = $ 6.626 \times 10^{-34} $ Js
$ c $ = speed of light = $ 3 \times 10^8 $ m/s
$ e $ = charge of electron = $ 1.6 \times 10^{-19} $ C
$ V $ = voltage = 20,000 V = $ 20 \times 10^3 $ V

Plug in the numbers:

$ \lambda_{min} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 20 \times 10^3} $

Solve the calculation:

$ \lambda_{min} = 0.062 \times 10^{-9}~\mathrm{m} $

We can also write this as:

$ \lambda_{min} = 0.062~\mathrm{nm} $

When light hits a photosensitive material, electrons are released. The energy of the incoming light (photon) is used to release these electrons and provide them with some extra energy.

Einstein’s photoelectric equation helps us understand this process:

$ E_{\max} = h \nu - \phi_0 $

This means: the maximum energy of the emitted electrons equals the energy of the incoming photon minus the work function ($\phi_0$), which is the energy needed to release the electron.

The stopping potential ($V_0$) is linked to this energy by:

$ eV_0 = h \nu - \phi_0 $

or

$ V_0 = \frac{h \nu}{e} - \frac{\phi_0}{e} $

First, use the given values for the first case:

For photon energy 3.1 eV and stopping potential 1.7 V:

$ V_0 = 3.1 - \frac{\phi_0}{e} $

$ 1.7 = 3.1 - \frac{\phi_0}{e} $

To find the work function divided by $e$:

$ \frac{\phi_0}{e} = 3.1 - 1.7 = 1.4 $

So, $ \frac{\phi_0}{e} = 1.4 \text{ volt} $

Now, use this value for the second case:

For photon energy 2.5 eV:

$ V_0 = 2.5 - \frac{\phi_0}{e} $

$ V_0 = 2.5 - 1.4 = 1.1 \text{ V} $

The maximum kinetic energy

$ \begin{aligned} & \mathrm{KE}_{\max }=E-\phi \\ & \mathrm{KE}_{\max }=4.5 \mathrm{eV}-3 \mathrm{eV} \\ & \mathrm{KE}_{\max }=1.5 \mathrm{eV} \end{aligned} $

The de-Broglie wavelength is,

$ \begin{aligned} & \lambda=\frac{12.27}{\sqrt{K E_{\max }}}\mathop {\rm{A}}\limits^{\rm{o}} \\ & \lambda=\frac{12.27}{\sqrt{1.5}} \mathop {\rm{A}}\limits^{\rm{o}} \Rightarrow \lambda \approx \frac{12.27}{1.2247} \mathop {\rm{A}}\limits^{\rm{o}} \\ & \lambda \approx 10.018 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

The de-Broglie wavelength associated with the photoelectrons is approximately $10.02 \mathop {\rm{A}}\limits^{\rm{o}}$.

According to Einstein's photoelectric equation,

$ \begin{aligned} & \left(E_K\right)_{\max }=E-\phi_0 \Rightarrow \phi_0=E-\left(E_K\right)_{\max } \\ & =8 \times 10^{-19}-\frac{p^2}{2 m} \\ & =8 \times 10^{-19}-\frac{\left(\frac{h}{\lambda}\right)^2}{2 m}=8 \times 10^{-19}-\frac{h^2}{2 m \lambda^2} \\ & =8 \times 10^{-19}-\frac{\left(6.62 \times 10^{-34}\right)^2}{2 \times 9.1 \times 10^{-31} \times\left(10 \times 10^{-10}\right)^2} \\ & =8 \times 10^{-19}-2.41 \times 10^{-19} \\ & =5.59 \times 10^{-19} \mathrm{~J} \\ & =\frac{5.59 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=3.49 \mathrm{eV} \approx 3.5 \mathrm{eV} \end{aligned} $

$ \begin{aligned} & \lambda_{\min }=\frac{h c}{\mathrm{eV}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 20 \times 10^3} \\ & =6.2 \times 10^{-11} \mathrm{~m} \\ & =0.62 \times 10^{-10} \mathrm{~m}=0.62 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

$ \begin{aligned} &\text { de-Broglie wavelength }\\ &\begin{aligned} & \lambda=\frac{h}{\sqrt{2 \mathrm{meV}}} \\ & =\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times \frac{200}{3}}} \\ & =1.5 \times 10^{-10} \mathrm{~m}=1.5 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { de-Broglie wavelength }\\ &\lambda=\frac{h}{\sqrt{3 m k_B T}} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad \lambda & \propto \frac{1}{\sqrt{T}} \\ \Rightarrow \quad \frac{\lambda_1}{\lambda_2} & =\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{(273+352)}{(273+127)}} \\ & =\sqrt{\frac{625}{400}}=\sqrt{\frac{25}{16}}=\frac{5}{4} \end{aligned} $

Step 1: Understanding Power

Power is how much energy is used or given out each second. So: $\text{Power} = \frac{\text{Energy}}{\text{Time}}$

Step 2: Energy of One Photon

The energy in one photon of light is found using: $E = h f$ where $h$ is Planck's constant and $f$ is the frequency.

Step 3: Total Energy from Many Photons

If $N$ photons come out in time $t$, the total energy is $N \times h f$.

Step 4: Linking Power and Number of Photons

So, $\text{Power} = \frac{\text{Total Energy}}{\text{Time}} = \frac{N \times h f}{t}$

Step 5: Average Number of Photons per Second

The number of photons per second is $n = \frac{N}{t} = \frac{P}{h f}$ Plug in the numbers: $n = \frac{33 \times 10^{-3}}{6.6 \times 10^{-34} \times 5 \times 10^{14}}$

Step 6: Final Calculation

This works out to: $n = 10 \times 10^{16}$ So, the laser gives off $10 \times 10^{16}$ photons every second.

Step 1: Formula for de-Broglie Wavelength

The de-Broglie wavelength is given by: $ \lambda = \frac{h}{\sqrt{2mk}} $

Step 2: Rearranging to Find Kinetic Energy

We can rearrange the formula to find kinetic energy $k$: $ \lambda^2 = \frac{h^2}{2mk} $ So, $ k = \frac{h^2}{2m\lambda^2} $

Step 3: Substitute the Values

Here,

• Planck's constant, $ h = 6.6 \times 10^{-34} $ Js
• Mass of electron, $ m = 9 \times 10^{-31} $ kg
• Wavelength, $ \lambda = 2 $ nm $ = 2 \times 10^{-9} $ m

Step 4: Calculation

Calculate the value: $ k = \frac{(6.6 \times 10^{-34})^2}{2 \times 9 \times 10^{-31} \times (2 \times 10^{-9})^2} $ This gives: $ k = 0.608 \times 10^{-19} \text{ J} $

Step 5: Convert Joules to Electron Volts

1 electron volt (eV) $ = 1.6 \times 10^{-19} $ J.
So, $ k = \frac{0.608 \times 10^{-19}}{1.6 \times 10^{-19}} \text{ eV} = 0.38 \text{ eV} $

The kinetic energy of the electron is about $ 0.38 $ eV.

$ \begin{aligned} &\text { de-Broglie wavelength }\\ &\begin{aligned} \lambda & =\frac{h}{\sqrt{2 m q V}} \\ \Rightarrow \quad \lambda & \propto \frac{1}{\sqrt{m q}} \\ \frac{\lambda_p}{\lambda_\alpha} & =\sqrt{\frac{m_\alpha}{m_p}} \sqrt{\frac{q_\alpha}{q_p}}=\sqrt{\frac{4}{1} \cdot \frac{2}{1}}=\frac{2 \sqrt{2}}{1} \\ \therefore \lambda_p: \lambda_\alpha & =2 \sqrt{2}: 1 \end{aligned} \end{aligned} $

Given:

Threshold wavelength,

$\lambda_0 = 6250 \, \text{Å} = 6250 \times 10^{-10} \, \text{m} = 6.25 \times 10^{-7} \, \text{m}$

Planck’s constant,

$h = 6.6 \times 10^{-34} \, \text{J·s}$

Speed of light,

$c = 3 \times 10^8 \, \text{m/s}$

Formula:

Work function $ \phi = h c / \lambda_0 $

Calculation:

$ \phi = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.25 \times 10^{-7}} $

$ \phi = \frac{19.8 \times 10^{-26}}{6.25 \times 10^{-7}} = 3.168 \times 10^{-19} \, \text{J} $

Converting to eV:

$1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$

$ \phi = \frac{3.168 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.98 \, \text{eV} $

✅ Answer: Option B — 1.98 eV

$ \begin{aligned} & \text { Since, } E=\frac{h c}{\lambda} \\ & \begin{aligned} \Rightarrow \lambda & =\frac{h c}{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{3 \times 1.6 \times 10^{-19}} \\ & =4.13 \times 10^{-7}=413 \times 10^{-9} \mathrm{~m} \\ & =413 \mathrm{~nm} \end{aligned} \end{aligned} $

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

when $\lambda^{\prime}=\lambda+0.25 \%$ of $\lambda=1.0025 \lambda$

So, $\quad 1.0025 \lambda=\frac{h}{p-p_0}$

$\Rightarrow \quad p-p_0=\frac{h}{1.0025}=\frac{p \lambda}{1.0025 \lambda}$

$\Rightarrow \quad \frac{p-p_0}{p}=\frac{1}{1.0025}$

$\Rightarrow \quad p=400 p_0$

$ \text { } \begin{aligned} E & =\frac{h c}{\lambda} \\ \Rightarrow \quad \lambda & =\frac{h c}{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{5 \times 1.6 \times 10^{-19}} \\ = & 248 \times 10^{-9} \mathrm{~m}=248 \mathrm{~nm} \end{aligned} $

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.