Dual Nature of Radiation

Given that the area of the metal plate $A = 1 \times 10^{-4} m^2$, and the intensity of the radiation

$I = 16 mW/m^2 = 16 \times 10^{-3} W/m^2$.

The power $P$ falling on the metal plate is given by the product of the intensity of the radiation and the area:

$P = IA = (16 \times 10^{-3} W/m^2)(1 \times 10^{-4} m^2) = 1.6 \times 10^{-6} W$.

The energy of each incident photon is given as 10 eV. Converting this to joules (since 1 eV = $1.6 \times 10^{-19} J$):

$E_{\text{photon}} = 10 eV = 10 \times 1.6 \times 10^{-19} J = 1.6 \times 10^{-18} J$.

The number of incident photons per second $n_{\text{photon}}$ is then given by the total power divided by the energy per photon:

$n_{\text{photon}} = P / E_{\text{photon}} = 1.6 \times 10^{-6} W / 1.6 \times 10^{-18} J = 10^{12} \text{ photons/s}$.

Given that only 10% of these photons actually produce photoelectrons, the number of photoelectrons produced per second $n_{\text{electron}}$ is:

$n_{\text{electron}} = 0.1 \times n_{\text{photon}} = 0.1 \times 10^{12} = 10^{11} \text{ electrons/s}$.

The work function of the metal is given as 5 eV, which is the energy required to remove an electron from the metal. Therefore, the maximum energy of the photoelectrons $E_{\text{max}}$ is given by the energy of the incident photons minus the work function:

$E_{\text{max}} = E_{\text{photon}} - \text{Work function} = 10 eV - 5 eV = 5 eV$.

We know that $E = {{hc} \over \lambda }$

So, for atom ${E_A} = {{hc} \over {{\lambda _A}}}$

And for neutron ${E_N} = {{hc} \over {{\lambda _N}}}$

Then, ${{{E_A}} \over {{E_N}}} = {{hc} \over {{\lambda _A}}} \times {{{\lambda _N}} \over {hc}} \Rightarrow {{{\lambda _N}} \over {{\lambda _A}}}$

Here, E_A is order of eV and E_N is order of MeV.

Therefore, ${{{\lambda _N}} \over {{\lambda _A}}} = {{{E_A}} \over {{E_N}}} = {{eV} \over {MeV}} = {{1eV} \over {{{10}^6}eV}}$

${\lambda _N} = {10^{ - 6}}\,{\lambda _A}$

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

From third Bohr's postulate, we have

$mvr = n{h \over {2\pi }}$

${h \over {mv}} = {{2\pi r} \over n} \Rightarrow \lambda = {{2\pi r} \over n}$

Since, $r = {a_0}{{{n^2}} \over Z}$, where a₀ is radius of Bohr's orbit having value (0.53) 1² $\mathop A\limits^o $ = 0.53 $\mathop A\limits^o $, therefore,

$\lambda = {{2\pi {a_0}{n^2}} \over {n\,.\,Z}} = {{2\pi {a_0}} \over Z}\,.\,n$

For Hydrogen Z = 1. Therefore,

${\lambda _G} = {{2\pi {a_0} \times 1} \over 1} = 2\pi {a_0}$ and ${\lambda _B} = {{2\pi {a_0} \times 3} \over 1} = 6\pi {a_0}$

Then, ${\lambda _B} = 3{\lambda _G}$

The power of the given laser light is expressed as

$P = {{nhc} \over {\lambda t}}$

from which the number of photons per second is given by

$n = P\left( {{{\lambda t} \over {hc}}} \right) = (5 \times {10^2}) \times \left[ {{{(660 \times {{10}^{ - 9}})(60 \times {{10}^{ - 3}})} \over {(6.6 \times {{10}^{ - 34}})(3 \times {{10}^8})}}} \right]$

$ = 100 \times {10^{18}} = {10^{20}}$

The given maximum velocity is v and frequency is n. We know that the kinetic energy is given by

$KE = {1 \over 2}m{v^2} = hn - \phi $

where h is Planck's constant and $\phi$ is the work function. Therefore, the kinetic energy of the incident light is

${E_1} = hn - \phi $ ..... (1)

When the frequency of the incident light is increased to 3n, then the kinetic energy is given by

${1 \over 2}mv_1^2 = 3hn - \phi $

$ \Rightarrow {E_2} = 3hn - \phi $ ..... (2)

Substituting $hn = {E_1} + \phi $ [from Eq. (1)] in Eq. (2), we get

${E_2} = 3({E_1} + \phi ) - \phi $

$ \Rightarrow {E_2} = 3{E_1} + 2\phi $

$ \Rightarrow {1 \over 2}mv_1^2 = 3 \times {1 \over 2}m{v^2} + 2\phi $

$ \Rightarrow v_1^2 = 3{v^2} + 2\phi \times {2 \over m}$

$ \Rightarrow v_1^2 = 3{v^2} + {{4\phi } \over m}$

Thus, the velocity is more than $\sqrt 3 v$.

Here, $\lambda$ is the wavelength of incident light and $\lambda$_d is the de Broglie wavelength of the fastest photoelectron. The fastest ejected photoelectron has the maximum kinetic energy which is given by

${K_{\max }} = {{hc} \over \lambda } - {\phi _0}$ .....(1)

The de Broglie wavelength of the photoelectron having kinetic energy ${K_{\max }} = {{{p^2}} \over {2m}}$ is given by

${\lambda _d} = {h \over p} = {h \over {\sqrt {2m{K_{\max }}} }}$ ..... (2)

where m is the mass of the electron and p is its linear momentum. Eliminate K_max from equations (1) and (2) to get

${{{h^2}} \over {2m\lambda _d^2}} = {{hc} \over \lambda } - {\phi _0}$ ...... (3)

Differentiate equation (3) to get

$( - 2){{{h^2}} \over {2m\lambda _d^3}}\Delta {\lambda _d} = ( - 1){{hc} \over {{\lambda ^2}}}\Delta \lambda $,

which gives

${{\Delta {\lambda _d}} \over {\Delta \lambda }} = {{mc} \over h}{{\lambda _d^3} \over {{\lambda ^2}}}$.

In photo-electric effect, the maximum kinetic energy of the photo-electron ejected at the cathode, is given by

${K_{\max ,\,c}} = {{hc} \over {{\lambda _{ph}}}} - \phi $,

where $\lambda$_ph is the wavelength of the incident light and $\phi$ is the work function of the material. The ejected photoelectrons are accelerated from the cathode to the anode by a potential V. Thus, the maximum kinetic energy of the photo-electron at the anode is

${K_{\max ,\,a}} = {K_{\max ,\,c}} + eV = {{hc} \over {{\lambda _{ph}}}} - \phi + eV$.

Now, minimum de-Broglie wavelength of electrons passing through anode,

${\lambda _e} = {h \over {\sqrt {2m\,{K_{\max ,\,a}}} }} = {h \over {\sqrt {2m\left( {{{hc} \over {{\lambda _{ph}}}} - \phi + eV} \right)} }}$ ...... (i)

D : For $V > > \phi e$

${\lambda _e} \approx {h \over {\sqrt {2meV} }} \propto {1 \over {\sqrt V }}$

$\therefore$ If V is made four times, $\lambda$_e becomes ${1 \over {\sqrt 4 }} = {1 \over 2}$, i.e. halved.

C : When $\phi$ and $\lambda$_ph increases, $\lambda$_e also increases.

A : From (i)

${{d{\lambda _e}} \over {dt}} = {{ - h} \over 2}{\left[ {2m\left( {{{hc} \over {{\lambda _{ph}}}} - \phi + eV} \right)} \right]^{ - 3/2}} \times 2m\left( {{{ - hc} \over {\lambda _{ph}^2}}} \right){{d{\lambda _{ph}}} \over {dt}}$

$ \Rightarrow {{d{\lambda _e}} \over {dt}} \ne {{d{\lambda _{ph}}} \over {dt}}$

I : $\lambda$_e does not depend on d.

Stopping potential (V₀) is given by

$e{V_0} = {{hc} \over \lambda } - \phi $

Graph between V₀ and $\lambda$ :

$e{V_0} + \phi = {{hc} \over \lambda }$

$(e{V_0} + \phi )\lambda = hc$

$(e{V_0} + \phi )\lambda $ = constant

Here, both e and $\phi$ are also constant. It represents a hyperbola.

For, V₀ = 0, $\lambda = {{constant} \over \phi } = $ constant

So correct option is (a).

Graph between V₀ and ${1 \over \lambda }$ :

${V_0} = \left( {{{hc} \over e}} \right)\left( {{1 \over \lambda }} \right) - \left( {{\phi \over e}} \right)$

It represents a straight line with slope $\left( {{{hc} \over e}} \right)$ and intercept $\left( { - {\phi \over e}} \right)$ on V₀ axis.

From Einstein's photoelectric equation,

kinetic energy of photoelectrons,

$K = {1 \over 2}m{v^2} = {{hc} \over \lambda } - \phi $ ..... (i)

For $\lambda$ = 248 nm, v = u₁

$\therefore$ ${K_1} = {1 \over 2}mu_1^2 = {{hc} \over {248}} - \phi $ ...... (ii)

For $\lambda$ = 310 nm, v = u₂

$\therefore$ ${K_2} = {1 \over 2}mu_2^2 = {{hc} \over {310}} - \phi $ ....... (iii)

Divide eqn. (ii) by eqn. (iii),

${{u_1^2} \over {u_2^2}} = {{{{hc} \over {248}} - \phi } \over {{{hc} \over {310}} - \phi }} = {{{{1240} \over {248}} - \phi } \over {{{1240} \over {310}} - \phi }}$

$ \Rightarrow {\left( {{2 \over 1}} \right)^2} = {{5 - \phi } \over {4 - \phi }}$

$ \Rightarrow 16 - 4\phi = 5 - \phi \Rightarrow 3\phi = 11$

or, $\phi$ = 3.67 eV $\approx$ 3.7 eV

To determine how the error in $d$ behaves as $\theta$ increases, let’s examine the given expression:

$2d\sin \theta = \lambda$

We can rearrange this to solve for $d$:

$d = \frac{\lambda}{2 \sin \theta}$

Given that the wavelength $\lambda$ is exactly known, any error in the measurement of $\theta$ (denoted as $\Delta \theta$) will propagate through to the error in $d$. To analyze this, we need to consider how the error propagates in the expression for $d$.

Assuming the error propagation formula, the differential of $d$ with respect to $\theta$ is:

$d = \frac{\lambda}{2 \sin \theta}$

Differentiating with respect to $\theta$:

$\frac{d(d)}{d(\theta)} = \frac{d}{d(\theta)} \left( \frac{\lambda}{2 \sin \theta} \right)$

Using the chain rule, we get the partial derivative:

$\frac{d(d)}{d(\theta)} = -\frac{\lambda \cos \theta}{2 (\sin \theta)^2}$

The absolute error in $d$, denoted as $\Delta d$, is given by:

$\Delta d = \left| \frac{d(d)}{d(\theta)} \Delta \theta \right| = \left| -\frac{\lambda \cos \theta}{2 (\sin \theta)^2} \right| \Delta \theta = \frac{\lambda \cos \theta}{2 (\sin \theta)^2} \Delta \theta$

Here, $\Delta \theta$ is the constant error in the measurement of $\theta$. From the equation above, we observe that $\Delta d$ depends on $\cos \theta / (\sin \theta)^2$ which is a function of $\theta$.

As $\theta$ increases from $0^\circ$ to $90^\circ$, the behavior is as follows:

• When $\theta$ is close to $0^\circ$, $\cos \theta \approx 1$ and $\sin \theta \approx 0$, thus $\left( \frac{\cos \theta}{(\sin \theta)^2} \right)$ becomes very large, indicating that $\Delta d$ is very large.
• As $\theta$ approaches $90^\circ$, $\cos \theta \approx 0$, and $\sin \theta \approx 1$, which makes $\left( \frac{\cos \theta}{(\sin \theta)^2} \right)$ approach zero, thus making $\Delta d$ very small.

Therefore, as $\theta$ increases from $0^\circ$ to $90^\circ$, the absolute error in $d$ decreases.

To analyze the fractional error in $d$ (denoted as $\frac{\Delta d}{d}$), we note:

$d = \frac{\lambda}{2 \sin \theta}$

So the fractional error is:

$\frac{\Delta d}{d} = \frac{\Delta d}{\frac{\lambda}{2 \sin \theta}} = \frac{\frac{\lambda \cos \theta}{2 (\sin \theta)^2} \Delta \theta}{\frac{\lambda}{2 \sin \theta}} = \frac{\cos \theta \Delta \theta}{\sin \theta}$

As $\theta$ increases from $0^\circ$ to $90^\circ$:

• Initially (at $\theta \approx 0^\circ$), $\cos \theta \approx 1$ and $\sin \theta \approx 0$, making the fractional error large.
• As $\theta$ continues to increase, $\cos \theta$ decreases and $\sin \theta$ increases, thus the fractional error $\frac{\cos \theta \Delta \theta}{\sin \theta}$ decreases.

Hence, the fractional error in $d$ also decreases as $\theta$ increases.

So the correct answer is:

The photoelectric effect equation, which relates the energy of the incident light to the kinetic energy of the ejected photoelectrons and the work function of the metal, is given by:

$E = K_{\max} + W$,

where

$E$ is the energy of the incident light,

$K_{\max}$ is the maximum kinetic energy of the photoelectrons, and

$W$ is the work function of the metal.

The energy of the incident light can be calculated using the formula $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the light. However, given that $hc = 1240$ eV⋅nm, we can simplify this to $E = \frac{1240}{\lambda}$.

Substituting the given values into this equation, we have:

$E = \frac{1240}{400} = 3.1$ eV.

We can then substitute these values into the photoelectric effect equation:

$3.1 \text{ eV} = 1.68 \text{ eV} + W$,

which simplifies to:

$W = 3.1 \text{ eV} - 1.68 \text{ eV} = 1.42$ eV.

Rounding to two decimal places, the work function of the metal is therefore approximately $1.42$ eV.

Thus, Option A: $1.41$ eV is the closest to the correct answer.

We have

$E = \left( {{{hc} \over \lambda }} \right)$ J $ \Rightarrow E = \left( {{{1240} \over {\lambda \,nm}}} \right)$ eV

Therefore,

$\lambda_1$ = 550 nm, $E_1$ = 2.25 eV

$\lambda_2$ = 450 nm, $E_2$ = 2.75 eV

$\lambda_3$ = 350 nm, $E_3$ = 3.5 eV

Also,

$\phi_p=2$ eV, all $\lambda$'s cause emissions.

$\phi_q=2.5$ eV, last two $\lambda$'s cause emissions.

$\phi_r=3$ eV, only the last $\lambda$ causes emissions.

That is, $I_p > I_q > I_r$.

We have,

$a = n\left( {{\lambda \over 2}} \right) \Rightarrow \lambda = {{2a} \over n}$

From de Broglie relation, we have

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

Therefore, ${{2a} \over n} = {h \over p} \Rightarrow p = {{nh} \over {2a}}$

Now, $E = {{{p^2}} \over {2m}} = {{{n^2}{h^2}} \over {8m{a^2}}} \Rightarrow E \propto {a^{ - 2}}$

For ground state, $n = 1$. Therefore,

${E_1} = {{{h^2}} \over {8m{a^2}}} = \left[ {{{{{(6.6 \times {{10}^{ - 34}})}^2}} \over {8 \times {{10}^{ - 30}} \times {{(6.6 \times {{10}^{ - 9}})}^2}}}} \right]$ J

Therefore, ${{{E_1}} \over e} = \left( {{{{E_1}} \over {1.6 \times {{10}^{ - 19}}}}} \right)$ meV = 8 meV

The cut-off wavelength is given by ${\lambda _{\min }} = {{hc} \over {eV}}$

The cut-off wavelength depends on the energy eV of the accelerated electrons and is independent of the atomic number of target. Thus, greater the accelerating voltage for electrons, higher will be kinetic energy it attains before striking the target, higher will be the frequency of X-rays and smaller will be wavelength.

The cut off wavelength is given by

$\lambda_{0}=\frac{h c}{e v}$ ..... (i)

According to de-Broglie equation

$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m e \mathrm{~V}}}$

$\Rightarrow \quad \lambda^{2}=\frac{h^{2}}{2 m e \mathrm{~V}}$

$\Rightarrow \quad \mathrm{V}=\frac{h^{2}}{2 m e \lambda^{2}}$ ..... (ii)

From (i) and (ii)

$\lambda_{0}=\frac{h c \times 2 m e \lambda^{2}}{e h^{2}}=2 m c \lambda^{2}$

Reason : Characteristics x-ray are produced due to transition of electron from one energy level to another. Similarly, the line in the hydrogen spectrum is obtained due to transition of electron from one energy level to another.

A $\to$ P, R

Reason : In photoelectric effect electron from the metal surface are emitted upon the incidence of light of appropriate freaquency.

Note : In $\beta$-decay, electrons are emitted from the nucleus of atom.

B $\to$ Q, S

Moseley gave a law which related frequency of emitted X-ray with the atomic number of the target material.

$\sqrt v=a(z-b)$

C $\to$ P

In photo electric effect, energy of photons of incident ray gets converted into K.E of emitted electron.

D $\to$ Q

Since, Potential increases

If we give the constant energy release we observe constant wavelength. So the statement 1 is true.

From Einstein equation,

${{hc} \over \lambda } = W + K.E$

W = work function (2.4) eV

If we given energy above 2.4 eV it will release photon.

Statement 2 is true.

Hint : ($\lambda_{ray}$ $\mu$ target material) does not depend on V$_0$ (accelerating potential)

Statement 1 is True

Statement 2 is True, it follows the energy conservatioin.

But statement 2 is not correct explanation of statement 1.

The energy of the incident radiation that causes photoelectrons to be emitted can be calculated using the stopping potential, V, via the equation

$E = eV,$

where e is the elementary charge.

The elementary charge, e, is approximately equal to $1.602 \times 10^{-19}$ C. So the energy of the incident radiation is

$E = 1.602 \times 10^{-19} \, \text{C} \times 5 \, \text{V} = 8.01 \times 10^{-19} \, \text{J}.$

We convert this to electron volts (eV) by using the conversion factor $1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}.$ So

$E = \frac{8.01 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} = 5 \, \text{eV}.$

This is less than the threshold energy of $6.2 \, \text{eV},$ which means that the incident radiation does not have enough energy to overcome the work function of the metal and thus cannot cause photoelectrons to be emitted.

The regions of the electromagnetic spectrum are generally classified by energy as follows:

• Infrared: Less than about $1.24 \, \text{eV}$
• Visible: Between about $1.24 \, \text{eV}$ and $3.1 \, \text{eV}$
• Ultraviolet: Between about $3.1 \, \text{eV}$ and $124 \, \text{eV}$
• X-rays: Greater than about $124 \, \text{eV}$

Therefore, the incident radiation falls in the ultraviolet region.