Dual Nature of Radiation

To solve this problem, we need to understand the relationship between the atomic number $ Z $, the characteristic X-ray wavelengths, and the cut-off wavelength in X-ray spectra.

The cut-off wavelength, also known as the minimum wavelength ($ \lambda_{\text{min}} $), corresponds to the maximum energy of the X-rays produced and can be related to the accelerating voltage used in the X-ray tube. The characteristic X-rays, such as the $ K_\alpha $-line, correspond to specific electron transitions in the atom and are dependent on the atomic number $ Z $ of the target material.

The wavelength of the $ K_\alpha $-line is given by Moseley's Law, which can be approximated as:

$ \lambda_{K_\alpha} \propto \frac{1}{(Z - 1)^2} $

The cut-off wavelength is given by the relationship between the energy of the incident electrons and the X-ray produced:

$ \lambda_{\text{min}} = \frac{hc}{eV} $

Now, given the ratio $ r $ for the first metal target with $ Z = 46 $ :

$ r = \frac{\lambda_{K_\alpha}}{\lambda_{\text{min}}} = 2 $

For the second metal target with $ Z = 41 $, let's denote the new ratio as $ r_2 $. We assume the same cutoff wavelength since the same electron beam is used.

The ratio for $ Z = 41 $ can be calculated using the relationship between the $ K_\alpha $-line wavelenghts:

$ \frac{\lambda_{K_\alpha (1)}}{\lambda_{K_\alpha (2)}} = \left( \frac{Z_2 - 1}{Z_1 - 1} \right)^2 $

Substituting $ Z_1 = 46 $ and $ Z_2 = 41 $:

$ \frac{\lambda_{K_\alpha (1)}}{\lambda_{K_\alpha (2)}} = \left( \frac{41 - 1}{46 - 1} \right)^2 = \left( \frac{40}{45} \right)^2 = \left( \frac{8}{9} \right)^2 = \frac{64}{81} $

Let $ \lambda_{K_\alpha (1)} = 2 \lambda_{\text{min}} $ because $ r = 2 $. Therefore:

$ \lambda_{K_\alpha (2)} = \lambda_{K_\alpha (1)} \cdot \frac{81}{64} = 2 \lambda_{\text{min}} \cdot \frac{81}{64} $

Thus, the new ratio $ r_2 $ is:

$ r_2 = \frac{\lambda_{K_\alpha (2)}}{\lambda_{\text{min}}} = 2 \cdot \frac{81}{64} = \frac{162}{64} = 2.53 $

So, the value of $ r $ for the second metal target with $ Z = 41 $ is:

Option A: 2.53

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

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

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

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.

$U(x) = \left\{ {\matrix{ {{E_0}} & {0 \le x \le 1} \cr 0 & {x > 1} \cr } } \right.$

$K.E = \left\{ {\matrix{ {2{E_0} - {E_0} = {E_0}} & {0 \le x \le 1} \cr {2{E_0} - 0 = 2{E_0}} & {x > 1} \cr } } \right.$

From $\lambda = {h \over {\sqrt {2mE} }}$

${\lambda _1} = {h \over {\sqrt {2m{E_0}} }}$ and ${\lambda _2} = {h \over {\sqrt {2m{E_0}} }} = {h \over {\sqrt {4m{E_0}} }}$

Now ${{{\lambda _1}} \over {{\lambda _2}}} = {{{h \over {\sqrt {2m{E_0}} }}} \over {{h \over {\sqrt {4m{E_0}} }}}} = {{\sqrt {4m{E_0}} } \over {\sqrt {2m{E_0}} }} = \sqrt {{{4m{E_0}} \over {2m{E_0}}}} \sqrt 2 $

${{{\lambda _1}} \over {{\lambda _2}}} = \sqrt 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.

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 energy of incident photon $\frac{h c}{\lambda}$ is equal to the sum of minimum energy required to emit the photoelectron (work function) and kinetic energy K of the ejected photoelectron.

The stopping potential V converts the kinetic energy K of ejected photoelectron into potential energy eV , there by stopping its motion.

Thus, $\frac{h c}{\lambda}=\phi+\mathrm{K}=\phi+\mathrm{eV}$

$ \Rightarrow \frac{h c}{\lambda}-\phi=\mathrm{eV} . $

i.e., $\mathrm{V}=\frac{h c}{e} \cdot \frac{1}{\lambda}-\frac{\phi}{e} \quad$ Slope, $m=\tan \theta=\frac{h c}{e}$

(on comparing with $y=m x-c$ )

At $\mathrm{V}=0, \phi=\frac{h c}{\lambda}$ which gives

$ \phi_1=0.001 h c, \phi_2=0.002 h c, \phi_3=0.004 h c, $

(For metal 1)           (metal 2)              (metal 3)

Thus, $\phi_1: \phi_2: \phi_3:: 1: 2: 4$.

Also, $\phi=\frac{h c}{\lambda_0}$, where $\lambda_0$ is threshold wavelength,

$\therefore$ The wavelength corresponding to $\phi_1, \phi_2, \lambda_3$ are

$ \begin{aligned} & \lambda_1=\frac{1}{0.001}=1000 \mathrm{~nm} \\ & \lambda_2=\frac{1}{0.002}=500 \mathrm{~nm} \\ & \lambda_3=\frac{1}{0.004}=250 \mathrm{~nm} \end{aligned} $

The wavelength of violet colour light is close to 400 nm , which is more than the threshold wavelength for metal 3 . Thus violet light can eject photoelectrons from from metal 2 but not from metal 3 .