Dual Nature of Radiation

The de Broglie wavelength is given by :

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

Since kinetic energy $\mathrm{K}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}$, we can write momentum as $\mathrm{p}=\sqrt{2 \mathrm{mK}}$.

Therefore:

$ \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}} $

Let $\mathrm{m}_{\mathrm{p}}$ be the mass of a proton or neutron.

Deuteron (d) consists of 1 proton and 1 neutron. The mass of deuteron is $\mathrm{m}_{\mathrm{d}} \approx 2 \mathrm{~m}_{\mathrm{p}}$

Kinetic Energy $\mathrm{K}_{\mathrm{d}}=\mathrm{E}$

Alpha particle $(\alpha)$ consists of 2 protons and 2 neutrons. The mass of alpha particle is $\mathrm{m}_\alpha \approx 4 \mathrm{~m}_{\mathrm{p}}$

Kinetic Energy $\mathrm{K}_\alpha=2 \mathrm{E}$

The ratio of wavelengths is :

$ \frac{\lambda_{\mathrm{d}}}{\lambda_\alpha}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{d}} \mathrm{~K}_{\mathrm{d}}}}}{\frac{\mathrm{~h}}{\sqrt{2 \mathrm{~m}_\alpha \mathrm{K}_\alpha}}}=\sqrt{\frac{\mathrm{m}_\alpha}{\mathrm{m}_{\mathrm{d}}}} \times \sqrt{\frac{\mathrm{K}_\alpha}{\mathrm{K}_{\mathrm{d}}}} $

Substituting the values:

$ \frac{\lambda_{\mathrm{d}}}{\lambda_\alpha}=\sqrt{\frac{4 \mathrm{~m}_{\mathrm{p}}}{2 \mathrm{~m}_{\mathrm{p}}}} \times \sqrt{\frac{2 \mathrm{E}}{\mathrm{E}}}=2 $

So, the ratio of de-Broglie wavelengths is $2: 1 \Rightarrow \frac{\mathrm{n}}{1}=\frac{2}{1} \Rightarrow \mathrm{n}=2$

Therefore, the value of $n$ is 2 . Hence, the correct answer is 2.

When a particle of charge $q$ is accelerated through a potential difference $V$, the work done on the particle is converted into its kinetic energy (K):

$ \mathrm{K}=\mathrm{qV} $

Given:

• Charge $\mathrm{q}=3 \times 10^{-19} \mathrm{C}$

• Potential $\mathrm{V}=1.21 \mathrm{~V}$

$ K=\left(3 \times 10^{-19}\right) \times(1.21)=3.63 \times 10^{-19} \mathrm{~J} $

The momentum ( p ) of a particle of mass m is related to its kinetic energy ( K ) by:

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

Substituting $\mathrm{K}=\mathrm{qV}$ :

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

The de-Broglie wavelength is given by:

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

Substituting the expression for momentum:

$ \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}} $

Substituting the given values:

$ \mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{~m}=6 \times 10^{-27} \mathrm{~kg}, \mathrm{q}=3 \times 10^{-19} \mathrm{C} \text { and } \mathrm{V}=1.21 \mathrm{~V} $

So, the de-Broglie wavelength associated with the particle is,

$ \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times\left(6 \times 10^{-27}\right) \times\left(3 \times 10^{-19}\right) \times 1.21}} $

$\Rightarrow $ $\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{36 \times 10^{-46} \times 1.21}}$

$\Rightarrow $ $\lambda=\frac{6.6 \times 10^{-34}}{6 \times 10^{-23} \times 1.1}$

$\Rightarrow $ $\lambda=\frac{6.6 \times 10^{-34}}{6.6 \times 10^{-23}}$

$\Rightarrow $ $\lambda=1 \times 10^{-11} \mathrm{~m}=\alpha \times 10^{-12} \mathrm{~m}$

$\Rightarrow $ $ \lambda=10 \times 10^{-12}=\alpha \times 10^{-12} \Rightarrow \alpha=10 $

Therefore, the value of $\alpha$ is 10 .

Hence, the correct answer is $\mathbf{1 0}$.

Let the momentum of $\mathrm{e}^{-}$at any time t is p and its de-broglie wavelength is $\lambda$.

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

$\begin{aligned} & \frac{\mathrm{dp}}{\mathrm{dt}}=\frac{-\mathrm{h}}{\lambda^2} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\ & \mathrm{ma}=\mathrm{F}=-\frac{\mathrm{h}}{\lambda} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \quad[\mathrm{~m}=\text { mass of } \mathrm{e}] \end{aligned}$

Where, -ve sign represents decrease in $\lambda$ with time

$\mathrm{ma}=\frac{-\mathrm{h}}{(\mathrm{~h} / \mathrm{p})^2} \frac{\mathrm{~d} \lambda}{\mathrm{dt}}$

$\begin{aligned} & \mathrm{a}=-\frac{\mathrm{p}^2}{\mathrm{mh}} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\ & \mathrm{a}=-\frac{\mathrm{mv}^2}{\mathrm{~h}} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\ & \frac{\mathrm{~d} \lambda}{\mathrm{dt}}=-\frac{\mathrm{ah}}{\mathrm{mv}^2}\quad\text{.... (1)} \end{aligned}$

here, $\mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}}=\frac{\mathrm{e}}{\mathrm{m}} \frac{\sigma}{2 \varepsilon_0}$

$\mathrm{a}=\frac{\sigma \mathrm{e}}{2 \mathrm{~m} \varepsilon_0}$

and $\mathrm{v}=\mathrm{u}+\mathrm{at}$

$\mathrm{v}=\mathrm{at}$

Substituting values of a \& v in equation (1)

$\begin{aligned} & \frac{\mathrm{d} \lambda}{\mathrm{dt}}=-\frac{2 \mathrm{~h} \varepsilon_0}{\sigma \mathrm{t}^2} \\ & \Rightarrow \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \propto \frac{1}{\mathrm{t}^2} \\ & \Rightarrow \mathrm{n}=2 \end{aligned}$

$\begin{aligned} &\text { Since power emitting by a source is given as }\\ &\begin{aligned} & =\frac{\text { Total energy emitted }}{\text { time }} \\ & =\frac{\left(E_1 \text { photon }\right) \times \text { Number of photons }(N)}{t} \\ & P_1=\left(E_1\right) n \end{aligned} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\left(\mathrm{E}_1\right) \mathrm{n}_1}{\left(\mathrm{E}_2\right) \mathrm{n}_2}=\frac{\left(\frac{\mathrm{hC}}{\lambda_1}\right) \mathrm{n}_1}{\left(\frac{\mathrm{hC}}{\lambda_2}\right) \mathrm{n}_2} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\left(\frac{\lambda_2}{\lambda_1}\right) \frac{\mathrm{n}_1}{\mathrm{n}_2} \end{aligned}\\ &\text { Substituting the given values }\\ &\begin{aligned} & 2=\left(\frac{300}{600}\right) \times \frac{2 \times 10^{15}}{\mathrm{n}_2} \\ & \mathrm{n}_2=\frac{1}{2} \times 10^{15}=5 \times 10^{14} \text { Photon } / \mathrm{sec} \end{aligned} \end{aligned}$

The Franck-Hertz experiment provides evidence for quantized energy levels within atoms. When atoms are excited by electrons with a specific kinetic energy, they can jump to higher energy levels. Upon returning to lower levels, they emit photons whose energies correspond to the difference between these levels. The first dip in the current-voltage graph for hydrogen, observed at $10.2 \mathrm{~V}$, corresponds to the energy required to excite a hydrogen atom to its first excitation level. The wavelength of the light emitted when the atom returns to its ground state can be calculated using the energy of the photon emitted.

To find the wavelength ($\lambda$) of light emitted, we use the relationship between energy ($E$), Planck's constant ($h$), the speed of light ($c$), and wavelength ($\lambda$), given in the equation form as $E = \frac{hc}{\lambda}$.

However, we are given $hc$ in electron volts per nanometer ($1245 \mathrm{~eV} \cdot \mathrm{nm}$), and the energy is also given in terms of voltage ($10.2 \mathrm{~V}$). First, we convert the energy into electron volts (eV) using the formula: $E = eV$, where $e$ is the charge of an electron ($1.6 \times 10^{-19} \mathrm{C}$).

The energy in electron volts can be directly calculated as:

$E = 10.2 \mathrm{~V} \cdot 1.6 \times 10^{-19} \mathrm{C/electron} = 10.2 \mathrm{~eV}$,

since $1 \mathrm{~V} \cdot 1 \mathrm{~C} = 1 \mathrm{~eV}$ by definition.

Next, using the energy-wavelength relationship and the given value for $hc$, the wavelength can be calculated as:

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

Substituting the given values yields:

$\lambda = \frac{1245 \mathrm{~eV} \cdot \mathrm{nm}}{10.2 \mathrm{~eV}}$

This simplifies to:

$\lambda = \frac{1245}{10.2} \mathrm{~nm}$

Calculating this gives:

$\lambda \approx 122.06 \mathrm{~nm}$.

Therefore, the wavelength of light emitted by the hydrogen atom when excited to the first excitation level is approximately $122.06 \mathrm{~nm}$.

The energy $E$ of a photon is related to its wavelength $\lambda$ by the formula:

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

where $h$ is Planck's constant and $c$ is the speed of light. In this problem, we are given that an atom absorbs a photon of wavelength $\lambda_1=500~\mathrm{nm}$ and emits another photon of wavelength $\lambda_2=600~\mathrm{nm}$. We can use the formula above to calculate the energy absorbed by the atom:

$\mathrm{Energy~absorbed}=E_1-E_2=\frac{hc}{\lambda_1}-\frac{hc}{\lambda_2}=hc\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)$

Substituting the given values for $h$ and $c$, we get:

$\mathrm{Energy~absorbed}=6.6\times10^{-34}~\mathrm{J\cdot s}\cdot3\times10^8~\mathrm{m/s}\cdot\left(\frac{1}{500\times10^{-9}~\mathrm{m}}-\frac{1}{600\times10^{-9}~\mathrm{m}}\right)$

Simplifying this expression, we get:

$\mathrm{Energy~absorbed}=6.6\times10^{-20}~\mathrm{J}$

We need to express this energy in electron volts (eV), which is a more convenient unit for atomic and molecular energies. To do this, we can divide the energy in joules by the charge of an electron:

$\mathrm{Energy~absorbed~in~eV}=\frac{6.6\times10^{-20}~\mathrm{J}}{1.6\times10^{-19}~\mathrm{C/eV}}=0.4125~\mathrm{eV}$

Finally, we can express the net energy absorbed in terms of the given value of $n$, as follows:

$n\times10^{-4}~\mathrm{eV}=0.4125~\mathrm{eV}$

Solving for $n$, we get:

$n=\frac{0.4125}{10^{-4}}=4125$

Therefore, the value of $n$ is $\boxed{4125}$.

When a monochromatic light is incident on hydrogen atoms in the ground state (n = 1), the hydrogen atoms can absorb energy and transition to higher energy levels. When the atoms return to lower energy levels, they emit radiation of different wavelengths corresponding to the energy differences between the energy levels.

The energy levels of the hydrogen atom are given by the formula:

$E_n = -\frac{13.6 \mathrm{~eV}}{n^2}$

where $E_n$ is the energy of the nth level and $n$ is the principal quantum number.

Since the hydrogen atoms emit radiation of six different wavelengths, there must be six different transitions from the excited states back to lower energy levels.

The six transitions correspond to the following energy level changes:

1. From n = 2 to n = 1
2. From n = 3 to n = 1
3. From n = 3 to n = 2
4. From n = 4 to n = 1
5. From n = 4 to n = 2
6. From n = 4 to n = 3

The highest energy level involved is n = 4. Therefore, the incident light must have a frequency high enough to excite the hydrogen atoms from the ground state (n = 1) to n = 4.

The energy difference between these levels is:

$\Delta E = E_4 - E_1 = \frac{13.6 \mathrm{~eV}}{4^2} - \frac{13.6 \mathrm{~eV}}{1^2} = -0.85 \mathrm{~eV} + 13.6 \mathrm{~eV} = 12.75 \mathrm{~eV}$

The frequency of the incident light is related to the energy difference by the equation:

$\Delta E = h \nu$

where $h$ is the Planck's constant and $\nu$ is the frequency.

Now, we can solve for the frequency:

$\nu = \frac{\Delta E}{h} = \frac{12.75 \mathrm{~eV}}{4.25 \times 10^{-15} \mathrm{~eVs}} = 3 \times 10^{15} \mathrm{~Hz}$

So, the value of $x$ is 3.

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

$ \Rightarrow {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {6630 \times {{10}^{ - 10}}}} - 6.63 \times {10^{ - 34}}{f_{th}} = 1.6 \times {10^{ - 19}} \times 0.4$

$ \Rightarrow {f_{th}} \simeq 35.11 \times {10^{13}}\,H$

Let us say that work function is $\phi$

$ \Rightarrow 2\phi = \phi + {1 \over 2}mv_1^2$ ...... (1)

and $5\phi = \phi + {1 \over 2}mv_2^2$ ..... (2)

From (1) and (2)

${{v_2^2} \over {v_1^2}} = {4 \over 1}$ or ${{{v_2}} \over {{v_1}}} = 2$

For photon ${\lambda _1} = {h \over P} = {{6.6 \times {{10}^{ - 34}}} \over {{{10}^{ - 27}}}}$

For particle ${\lambda _2} = {h \over {mv}} = {{6.6 \times {{10}^{ - 34}}} \over {9.1 \times {{10}^{ - 31}} \times {{10}^6}}}$

$\therefore$ ${{{\lambda _1}} \over {{\lambda _2}}} = 910$

By photoelectric equation

${{hc} \over \lambda } - \phi = {k_{\max }}$

${k_{\max }} = {{1240} \over {500}} - 1.25 \approx 1.25$

$r = {{\sqrt {2mk} } \over {eB}}$

$B = {{\sqrt {2mk} } \over {er}}$

$ = 125 \times {10^{ - 7}}T$

$KE = {{hc} \over \lambda } - \phi hc$

$e(3{V_0}) = {{hc} \over \lambda } - \phi $ ..... (i)

$e{V_0} = {{hc} \over {2\lambda }} - \phi $ ..... (ii)

Using (i) & (ii)

$\phi = {{hc} \over {4{\lambda _0}}} = {{hc} \over {{\lambda _t}}}$

${\lambda _t} = 4{\lambda _0}$

$K{E_{\max }} = hv - \phi $

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

$v = \sqrt {{{2(hv - \phi )} \over m}} $

Given $h{v_1} = 2\phi $

$h{v_2} = 10\phi $

$ \therefore $ ${{{v_1}} \over {{v_2}}} = \sqrt {{{h{v_1} - \phi } \over {h{v_2} - \phi }}} $

${{{v_1}} \over {{v_2}}} = \sqrt {{{2\phi - \phi } \over {10\phi - \phi }}} = {1 \over 3}$ = ${x \over y}$

$ \therefore $ x = 1

Given,

E₁ = 4 eV

E₂ = 2.5 eV

and ${{{{\left( {{V_1}} \right)}_{\max }}} \over {{{\left( {{V_2}} \right)}_{\max }}}} = 2$

${{{1 \over 2}m{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{1 \over 2}m{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \right)}^2}}} = {{{E_1} - {\phi _0}} \over {{E_2} - {\phi _0}}}$

$ \Rightarrow $ ${{{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \right)}^2}}} = {{4 - {\phi _0}} \over {2.5 - {\phi _0}}}$

$ \Rightarrow $ ${\left( 2 \right)^2} = {{4 - {\phi _0}} \over {2.5 - {\phi _0}}}$

$ \Rightarrow $ 10 - 4$\phi $₀ = 4 - $\phi $₀ $ \Rightarrow $ 3$\phi $₀ = 6

$ \Rightarrow $ $\phi $₀ = 2

$ \therefore $ Work function($\phi $) of the metal = 2 eV

Given d = 1 $\mathop A\limits^o $

For first maxima, $\theta $ = 60^o

$ \therefore $ $\theta $₁ = 90 - ${\theta \over 2}$

= $90 - {{60} \over 2}$ = 60^o

and $2d\sin \theta = \lambda = {h \over {\sqrt {2mE} }}$

$ \Rightarrow $ $2 \times {10^{ - 10}} \times {{\sqrt 3 } \over 2} = {{6.6 \times {{10}^{ - 34}}} \over {\sqrt {2mE} }}$

$ \Rightarrow $ $E = {1 \over 2} \times {{6.64 \times {{10}^{ - 48}}} \over {9.1 \times {{10}^{ - 31}} \times 3 \times 1.6 \times {{10}^{ - 19}}}} = 50.47$

${{hc} \over \lambda } - \phi $ = eV .....(i)

${{hc} \over {3\lambda }} - \phi = {{eV} \over 4}$ .....(ii)

From (i) and (ii),

${{hc} \over {3\lambda }} - \phi =$ ${{hc} \over {4\lambda }} - {\phi \over 4}$

$ \Rightarrow $ ${{hc} \over \lambda }\left( {{1 \over 3} - {1 \over 4}} \right) = {{3\phi } \over 4}$

$ \Rightarrow $ ${{hc} \over {9\lambda }} = \phi $

Also, $\phi $ = ${{hc} \over {{\lambda _0}}}$

$ \therefore $ ${{hc} \over {9\lambda }} = {{hc} \over {{\lambda _0}}}$

$ \Rightarrow $ $\phi $ = 9$\lambda $

So, n = 9

Energy of photon = ${{1240} \over {310}}$ = 4 eV

Energy is greater than work function so photoelectric effect will take place.

Number of photons =

Intensity

Energy of one photon

= ${{6.4 \times {{10}^{ - 5}}} \over {4 \times 1.6 \times {{10}^{ - 19}}}}$ = 10¹⁴

Total number of electrons = ${{{{10}^{14}}} \over {{{10}^3}}}$ = 10¹¹

Ionization of H (ground state at –13.6 eV)

• Photon of energy $E_1=h\nu_1$ ionizes H and leaves the electron with 10 eV.

• Energy conservation (neglecting proton recoil):

$ E_1 = 13.6\;\text{eV} \;+\;10\;\text{eV} =23.6\;\text{eV}. $

Formation of positronium (Ps)

• Positron initially at rest, electron arrives with 10 eV.

• Ps ground‐state binding energy is 6.8 eV (half of hydrogen’s 13.6 eV).

• The newly formed Ps moves with 5 eV of center‐of‐mass (COM) kinetic energy and emits a photon of energy $E_2=h\nu_2$.

• Writing energy conservation (zero energy set at free e⁺ + e⁻ at rest):

$ \underbrace{10\;\text{eV}}_{\text{initial KE}} =\;\underbrace{(-6.8\;\text{eV})}_{\text{binding}} \;+\;\underbrace{5\;\text{eV}}_{\text{COM KE}} \;+\;E_2 $

so

$ E_2 = 10 -5 +6.8 = 11.8\;\text{eV}. $

Difference between the two photon energies

$ E_1 - E_2 = 23.6\;\text{eV} \;-\;11.8\;\text{eV} = 11.8\;\text{eV}. $

Answer: The difference $h\nu_1 - h\nu_2$ is 11.8 eV.

Here, we will first find the de Broglie wavelength of the electron in the $n=3$ orbit of a hydrogen-like atom with atomic number $Z$. Then we will find the de Broglie wavelength of the Neutron having thermal energy $k_B T$. Thereafter, we will equate both the wavelengths as they are the same (given)

de Broglie wavelength of the electron:

Radius in the $n$-th orbit,

$ \begin{aligned} & r_n=\frac{n^2 a_0}{Z} \\ & n=3 \text { So } r=\frac{3^2 a_0}{Z}=\frac{9 a_0}{Z} \end{aligned} $

Using Bohr quantization :

$ \begin{aligned} & m_e v r=n \hbar \\ \Rightarrow & m_e v \cdot \frac{n^2 a_0}{Z}= \frac{nh}{2 \pi} \\ \Rightarrow & p=m_e v=\frac{\hbar h}{2 \pi} \cdot \frac{Z}{n^2 a_0}=\frac{Z h}{2 \pi n a_0} \\ & n=3 \text { so } p=\frac{Z h}{6 \pi a_0} \end{aligned} $

de Broglie wavelength, $\lambda_e=\frac{h}{p}=\frac{h}{\frac{2 h}{6 \pi q_0}}$

$ \begin{aligned} &\Rightarrow \lambda_e=\frac{6 \pi a_0}{Z}\\ &\text { Neutron's de Broglie wavelength : }\\ &E=k_B T \text { (given) } \end{aligned} $

$ \begin{aligned} &\begin{aligned} p & =\sqrt{2 m_N E}=\sqrt{2 m_N k_B T} \\ \lambda_n & =\frac{h}{p}=\frac{h}{\sqrt{2 m_N k_B T}} \end{aligned}\\ &\text { Now, equating the wavelengths, }\\ &\begin{aligned} & \frac{6 \pi a_0}{Z}=\frac{h}{\sqrt{2 m_N k_B T}} \\ \Rightarrow & \sqrt{2 m_N k_B T}=\frac{Z h}{6 \pi a_0} \\ \Rightarrow & 2 m_N k_B T=\frac{Z^2 h^2}{36 \pi^2 a_0^2} \\ \Rightarrow & T=\frac{Z^2 h^2}{72 m_N k_B \pi^2 a_0^2} \\ & T=\frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B} \text { (given) } \end{aligned} \end{aligned} $

$ \text { So, } \alpha=72 $

To find the atomic number $ Z $ for the hydrogen-like atom emitting photons that cause photoelectrons to eject from the metal surface with a maximum kinetic energy of $ 1.95 \mathrm{eV} $, we need to make use of the concept of energy transitions in atoms, as well as the photoelectric effect equation. We are given the photoelectric threshold wavelength for the metal and the constants $ h c $ and $ R h c $.

First, let's calculate the energy of the photon emitted during the transition from level $ n=4 $ to level $ n=3 $ in the hydrogen-like atom. The energy of a photon emitted when an electron transitions between two levels in a hydrogen-like atom is given by :

$ \Delta E = Z^2 R h c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $

where:

• $ Z $ is the atomic number.
• $ R h c $ is the ionization energy of hydrogen (13.6 eV).
• $ n_1 $ and $ n_2 $ are the principal quantum numbers of the initial and final energy levels, respectively.

For the transition from $ n_1 = 4 $ to $ n_2 = 3 $, the energy of the photon is:

$ \Delta E = Z^2 \cdot 13.6 \cdot \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \mathrm{eV} $

$ \Delta E = Z^2 \cdot 13.6 \cdot \left( \frac{1}{9} - \frac{1}{16} \right) \mathrm{eV} $

$ \Delta E = Z^2 \cdot 13.6 \cdot \left( \frac{16 - 9}{144} \right) \mathrm{eV} $

$ \Delta E = Z^2 \cdot 13.6 \cdot \frac{7}{144} \mathrm{eV} $

Next, using the photoelectric effect equation, the maximum kinetic energy (K.E.) of the ejected photoelectrons is equal to the energy of the incident photon minus the work function ($ \phi $). The work function can be calculated using the given threshold wavelength :

$ \phi = \frac{h c}{\lambda} $

Given: $ \lambda = 310 \mathrm{nm} $ $ h c = 1240 \mathrm{eV} \cdot \mathrm{nm} $

So the work function is:

$ \phi = \frac{1240}{310} \mathrm{eV} $

$ \phi = 4 \mathrm{eV} $

The maximum kinetic energy is given by :

$ K.E. = \Delta E - \phi $

We are given $ K.E. = 1.95 \mathrm{eV} $, hence :

$ 1.95 = Z^2 \cdot 13.6 \cdot \frac{7}{144} - 4 $

$ \Rightarrow $ $ Z^2 \cdot 13.6 \cdot \frac{7}{144} = 1.95 + 4 $

$ \Rightarrow $ $ Z^2 = \frac{1.95 + 4}{13.6 \cdot \frac{7}{144}} $

$ \Rightarrow $ $ Z^2 = \frac{5.95 \times 144}{13.6 \times 7} $

$ \Rightarrow $ $ Z^2 = \frac{856.8}{95.2} $

$ \Rightarrow $ $ Z^2 = 9 $

$ \Rightarrow $ $ Z = \sqrt{9} $

$ \Rightarrow $ $ Z = 3 $

E_P = 2E_Q = 2E_R = E(assume)

For metal P and Q,

E₁ - 4 = E_P = E ……..(1)

E₁ - 4.5 = E_A = E/2 ……..(2)

Subtracting equation (2) from (1), we get

E/2 = 0.5

For metal R

E₂ - 5.5 = E_R = E/2 = 0.5

$ \Rightarrow $ E₂ = 6

Momentum transferred on mirror = ${{2Nh} \over \lambda }$

${{2Nh} \over \lambda } = M{V_{(mean\,position)}}$

${V_{(mean\,position)}} = \Omega A$ (where, A = 1 $\mu $m)

${{2Nh} \over \lambda }$$ = M\Omega A$

(where $\lambda $ = 8$\pi $ $ \times $ 10^{$ - $6})

$N = {{M\Omega ({{10}^{ - 6}})\lambda } \over {2h}}$

$ = {{M\Omega 8\pi \times {{10}^{ - 6}} \times {{10}^{ - 6}}} \over {2h}}$

$N = {{4\pi M\Omega } \over h} \times {10^{ - 12}}$

$ = {10^{ - 24}} \times {10^{ - 12}}$

$N = 1 \times {10^{ - 12}} \Rightarrow x = 1$

The frequency of incident light is just above the threshold frequency. Hence, the energy of each photon is equal to the work function (E_p = $\phi$ = 6.25 eV) and the kinetic energy of emitted photo-electron is zero (K_e0 = 0). The energy incident per second on the cathode is incident power P = 200 W. Thus, number of photons incident per second is

N_p = P/E_p = P/$\phi$.

The photo-electron emission efficiency is 100%. Thus, number of photo-electron emitted per second is equal to the number of photons incident per second i.e., N_e = N_p. These photo-electrons are accelerated by a potential difference V = 500 V. Thus, gain in potential energy of each photo-electron is $\Delta$U = eV = 500 eV. By conservation of energy, kinetic energy of the photo-electron when it reaches the anode is

K_e = K_e0 + $\Delta$U = 500 eV.

The linear momentum of photo-electron of mass m_e and kinetic energy K_e is given by

${p_e} = \sqrt {2{m_e}{K_e}} $.

The photo-electron transfer its entire linear momentum to the anode (absorbed by the anode). Thus, gain in linear momentum of the anode by absorbing one photo-electron is

$\Delta$p_a = p_e.

The force on the anode is equal to the increase in its linear momentum per second. Since N_e photo-electrons strikes the anode per second, the force acting on the anode is given by

${F_a} = {N_e}\Delta {p_a} = {N_e}{p_e}$ ($\because$ $\Delta {p_a} = {p_e}$)

$ = {N_p}{p_e} = (P/\phi ){p_e}$ ($\because$ ${N_e} = {N_p} = P/\phi $)

$ = (P/\phi )\sqrt {2{m_e}{K_e}} $ ($\because$ ${p_e} = \sqrt {2{m_e}{K_e}} $)

$ = (P/\phi )\sqrt {2{m_e}eV} $ ($\because$ ${K_e} = eV$

$ = 2.4 \times {10^{ - 4}}N$.

Angular momentum $mvr = {{nh} \over {2\pi }}$ where $r = 3{a_0}$ where $n = 3$, that is, electron in $L{i^{2 + }}$ is in second excited state

$\lambda = {h \over {mv}} = p\pi {a_0}$

$ \Rightarrow n = p\pi (mv{a_0}) = p\pi \left( {{{mvr} \over 3}} \right) = {{p\pi } \over 3}\left( {{{3h} \over {2\pi }}} \right) = {{ph} \over 2}$

Therefore, $p = 2$.

We have,

$V = {{hf} \over e} - {\phi \over e}$

Slope is h/e.

Slope is the same for both silver and sodium.

Therefore, the ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is 1 : 1.

Let initial and final kinetic energies of the proton be K_i and K_f and corresponding potential energies be U_i and U_f. When proton is far away from the nucleus (r $\to$ $\infty$), its potential energy is

${U_i} = \mathop {\lim }\limits_{r \to \infty } {1 \over {4\pi {\varepsilon _0}}}{{120{e^2}} \over r} = 0$.

At closest distance, the proton comes to rest momentarily, giving K_f = 0. The potential energy at closest distance is

${U_f} = {1 \over {4\pi {\varepsilon _0}}}{{120{e^2}} \over a}$,

where a is the distance of closest approach. Since electrostatic force is conservative, total energy is conserved i.e., ${K_i} + {U_i} = {K_f} + {U_f}$. Substitute the values to get

${K_i} = {U_f} = {1 \over {4\pi {\varepsilon _0}}}{{120{e^2}} \over a}$

The de-Broglie wavelength of the proton is given by

${\lambda _i} = {h \over {{p_i}}} = {h \over {\sqrt {2{m_p}{K_i}} }} = {h \over e}\sqrt {{{4\pi {\varepsilon _0}a} \over {240{m_p}}}} $

$ = 4.2 \times {10^{ - 15}}{\left( {{{10 \times {{10}^{ - 15}}} \over {9 \times {{10}^9} \times 240 \times (5/3) \times {{10}^{ - 27}}}}} \right)^{1/2}}$

= 7 fm.

Activity $A = \lambda N$, where $\lambda$ is decay constant and N is number of particles present. Therefore,

$N = {A \over \lambda } = A\tau $

where $\tau$ = 1 / $\lambda$ is the mean life of the sample. The mass of the sample is

$M = mN = mA\tau $

where m is mass of an atom. Therefore, the mass of the radioactive sample is

M = 10^$-$25 $\times$ 10¹⁰ $\times$ 10⁹ = 10^$-$6 kg = 1 mg

The silver sphere gets positively charged due to emission of photoelectrons. This positively charged sphere attracts (binds) the emitted photoelectrons. The emitted photoelectrons cannot escape if their kinetic energies (hc/$\lambda$ $-$ $\phi$) are less than or equal to their potential energies $\left( {{1 \over {4\pi {\varepsilon _0}}}{{n{e^2}} \over r}} \right)$. Thus, in limiting case,

${{hc} \over \lambda } - \phi = {1 \over {4\pi {\varepsilon _0}}}{{n{e^2}} \over r}$ ..... (1)

Substitute the values of various parameters in equation (1),

${{1242} \over {200}} - 4.7 = {{n(9 \times {{10}^9})(1.6 \times {{10}^{ - 19}})} \over {{{10}^{ - 2}}}}$,

to get n = 1.04 $\times$ 10⁷. [We have used hc = 1242 eV-nm.]

The de Broglie wavelength of a particle with momentum p is given by

$\lambda$ = h/p.

The momentum and kinetic energy of a particle of mass m are related by

$p = \sqrt {2mK} $.

The kinetic energy of a charge q, accelerated through potential V, is given by K = qV. Thus,

$\lambda = h/\sqrt {2mK} = h/\sqrt {2mqV} $,

which gives

${{{\lambda _p}} \over {{\lambda _\alpha }}} = \sqrt {{{2{m_\alpha }{q_\alpha }V} \over {2{m_p}{q_p}V}}} = \sqrt {{{2\,.\,4u\,.\,2e\,.\,100} \over {2\,.\,1u\,.\,1e\,.\,100}}} $

$ = \sqrt 8 = 2.8 \approx 3$