Dual Nature of Radiation

The photoelectric effect occurs when light (or more generally, electromagnetic radiation) incident on a metallic surface causes the ejection of electrons from the surface. The energy of the incident photons must be greater than the work function of the metal (denoted by $\phi_0$) for the electrons to be ejected.

The energy of the ejected electrons can be expressed as the difference between the energy of the incident photons and the work function of the metal. The maximum kinetic energy of the ejected electrons is given by:

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

where $h$ is the Planck's constant, $\nu$ is the frequency of the incident light, and $\phi_0$ is the work function of the metal.

We can also write the maximum kinetic energy in terms of the stopping potential ($V_0$) and the elementary charge ($e$) of an electron:

$K_{max} = eV_0$

Equating these two expressions for the maximum kinetic energy, we get:

$eV_0 = h\nu - \phi_0$

We can express the frequency $\nu$ in terms of the speed of light $c$ and the wavelength $\lambda$:

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

Substituting this expression for frequency in the equation, we get:

$eV_0 = h\frac{c}{\lambda} - \phi_0$

Now, we have two cases:

1. The stopping potential is $V_0$, and the wavelength is $\lambda$.
2. The stopping potential is $\frac{V_0}{4}$, and the wavelength is $2\lambda$.

For the first case, we use the photoelectric effect equation as is:

$eV_0 = \frac{hc}{\lambda} - \phi_0$

For the second case, we replace $V_0$ with $\frac{V_0}{4}$ and $\lambda$ with $2\lambda$:

$\frac{eV_0}{4} = \frac{hc}{2\lambda} - \phi_0$

Now we have two equations:

(1) $eV_0 = \frac{hc}{\lambda} - \phi_0$

(2) $\frac{eV_0}{4} = \frac{hc}{2\lambda} - \phi_0$

We can rewrite equation (1) as:

$\frac{eV_0}{4} = \frac{hc}{4\lambda} - \frac{\phi_0}{4}$

Now we can equate the two expressions for $\frac{eV_0}{4}$:

$\frac{hc}{4\lambda} - \frac{\phi_0}{4} = \frac{hc}{2\lambda} - \phi_0$

Now we isolate the terms containing $\phi_0$:

$\frac{3\phi_0}{4} = \frac{hc}{4\lambda} - \frac{hc}{2\lambda}$

$\frac{3\phi_0}{4} = \frac{hc}{4\lambda}$

Now we can write the work function $\phi_0$ in terms of the threshold wavelength $\lambda_0$:

$\phi_0 = \frac{hc}{\lambda_0}$

Substituting the expression for the work function $\phi_0$ in terms of the threshold wavelength $\lambda_0$ into the previous equation, we get:

$\frac{3}{4} \frac{hc}{\lambda_0} = \frac{hc}{4\lambda}$

Now we can cancel out the common terms $hc$ from both sides:

$\frac{3}{4\lambda_0} = \frac{1}{4\lambda}$

Next, we can cross-multiply to solve for $\lambda_0$:

$3\lambda = \lambda_0$

Thus, the threshold wavelength for this metallic surface is $3\lambda$.

The de Broglie wavelength of a particle is given by:

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

where h is Planck's constant and p is the momentum of the particle. The momentum of a gas molecule can be related to its kinetic energy (which is related to the temperature of the gas) by:

$p = \sqrt{2mK}$

where m is the mass of the molecule and K is the kinetic energy of the molecule.

At a given temperature T, the average kinetic energy of a molecule in a gas is given by:

$K = \frac{3}{2} kT$

where k is Boltzmann's constant.

Therefore, the de Broglie wavelength of a molecule in a gas is given by:

$\lambda = \frac{h}{\sqrt{2m(3/2)kT}} = \frac{h}{\sqrt{3mkT}}$

If the temperature of the gas is increased from T = 300 K to T = 600 K, the new de Broglie wavelength becomes:

$\lambda' = \frac{h}{\sqrt{3mk(2T)}} = \frac{h}{\sqrt{2} \sqrt{3mkT}} = \frac{1}{\sqrt{2}} \lambda$

So, the de Broglie wavelength of the gas molecule decreases by a factor of $\sqrt{2}$ when the temperature of the gas is doubled.

Therefore, the correct answer is $\lambda' = \frac{1}{\sqrt{2}} \lambda_1$

The photoelectric effect is the phenomenon of emission of electrons (or photoelectrons) from the surface of a metal when it is illuminated by light of sufficient energy. The observations from the photoelectric effect led to the development of quantum theory.

According to the principles of the photoelectric effect:

A. The photocurrent (number of photoelectrons ejected per unit time) is indeed proportional to the intensity of the incident radiation. More intense light means more photons hitting the surface and thus more electrons being ejected.

B. The maximum kinetic energy of the photoelectrons does not depend on the intensity of the incident light but rather on its frequency. Increasing the intensity of light increases the number of photoelectrons (current) but does not increase their maximum kinetic energy.

C. The maximum kinetic energy of the photoelectrons does indeed depend on the frequency of the incident light. If the frequency of the incident light is below a certain threshold frequency specific to the metal, no photoelectrons are emitted regardless of the intensity of the light. Above this threshold, the maximum kinetic energy of the photoelectrons increases linearly with the frequency of the light.

D. The emission of photoelectrons does not require a minimum threshold intensity of incident radiation, but rather a minimum threshold frequency.

E. The maximum kinetic energy of the photoelectrons is not independent of the frequency of the incident light, but rather depends on it.

Therefore, only statements A and C are correct.

The de Broglie wavelength of a particle is given by the formula:

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

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

If the de Broglie wavelengths of the proton and electron are the same, then:

$\frac{h}{p_p} = \frac{h}{p_e}$

where $p_p$ and $p_e$ are the momenta of the proton and electron, respectively.

Solving this equation for the ratio of their momenta gives:

$\frac{p_p}{p_e} = 1$

So, the ratio of their momenta is 1:1

We are given the work functions of Aluminium and Gold as $4.1 ~\mathrm{eV}$ and $5.1 ~\mathrm{eV}$, respectively.

The stopping potential ($V_s$) is related to the frequency ($f$) of the incident light by the equation:

$eV_s = h(f - f_0)$

Where $e$ is the charge of an electron, $h$ is Planck's constant, and $f_0$ is the threshold frequency. The threshold frequency is related to the work function ($\phi$) by:

$\phi = hf_0$

So, we can write the equation for stopping potential as:

$V_s = \frac{h}{e}(f - \frac{\phi}{h})$

This equation represents a straight line with slope $\frac{h}{e}$. Therefore, the slope of the stopping potential versus frequency plot is the same for both Aluminium and Gold. The ratio of the slopes is:

$\frac{\text{Slope for Gold}}{\text{Slope for Aluminium}} = \frac{\frac{h}{e}}{\frac{h}{e}} = 1$

So , the ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is 1.

The de Broglie wavelength of a particle is given by the equation:

$ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}, $

where:

• $h$ is Planck's constant,
• $p$ is the momentum of the particle,
• $m$ is the mass of the particle, and
• $K$ is the kinetic energy of the particle.

This equation shows that the de Broglie wavelength of a particle is inversely proportional to the square root of its mass and its kinetic energy. This means that the particle with the smallest mass and kinetic energy will have the largest de Broglie wavelength.

Given that the kinetic energies of the electron, alpha particle, and proton are $4K$, $2K$, and $K$, respectively, and knowing that the mass of the electron is less than the mass of the proton and the mass of the proton is less than the mass of the alpha particle, we can infer that the de Broglie wavelength of the electron is greater than the de Broglie wavelength of the proton, which in turn is greater than the de Broglie wavelength of the alpha particle.

Therefore, the correct answer is $\lambda_{\alpha} < \lambda_{p} < \lambda_{e}$.

The assertion A and the reason R are both correct statements, and the reason R is the correct explanation of the assertion A.

Explanation :

The assertion A states that the beam of electrons exhibit wave nature and show interference and diffraction. This statement is correct because electrons exhibit both particle-like and wave-like behavior. When electrons are accelerated to high speeds, they have a wavelength associated with them, and this wavelength can interfere and diffract just like any other wave.

The reason R states that the wave nature of electrons was experimentally verified by Davisson Germer. This statement is also correct because Davisson and Germer performed an experiment in 1927 where they observed diffraction patterns in a beam of electrons that were scattered off a nickel crystal. This observation provided strong evidence for the wave nature of electrons.

Therefore, both assertion A and reason R are correct statements, and reason R is the correct explanation of assertion A.

$P = \sqrt {2\,eVm} $

$\lambda = \left( {{h \over {{P_1}}}} \right)$ ..... (i)

${{3\lambda } \over 2} = {h \over {{P_2}}}$ ..... (ii)

Dividing (i) by (ii)

$ \Rightarrow {2 \over 3} = \left( {{{{P_2}} \over {{P_1}}}} \right) = \sqrt {{{{v_2}} \over {{v_1}}}} $

$ \Rightarrow {4 \over 9} = \left( {{{{v_2}} \over {{v_1}}}} \right)$

${{{v_1}} \over {{v_2}}} = \left( {{9 \over 4}} \right)$

Assuming the small object as photon.

Momentum $(p)=\frac{E}{C}$

$=\frac{20\times10^{-3}\times300\times10^{-9}}{3\times10^8}$

$=2\times10^{-17}$ kg m/s

The similar materials will have same stopping potential and the different materials will have different stopping potential whatever may be the intensity of incident radiation.

So, curve $(1,2)$ and $(3,4)$ represent same materials.

So, option (d) is the correct option.

Given, Radiation's wavelength, $\lambda=400 \mathrm{~nm}$ Work function, $\phi=2.2 \mathrm{eV}$ Energy of incident photon is given by

$ E=\frac{1240}{\lambda(\mathrm{~nm})} \mathrm{eV}=\frac{1240}{400}=3.1 \mathrm{eV} $

Maximum kinetic energy of ejected electron

$ \mathrm{KE}=E-\phi=3.1-2.2=0.9 \mathrm{eV} $

Potential is energy/unit charge

$ \begin{aligned} V & =\frac{0.9 \mathrm{eV}}{e} \\ \Rightarrow \quad V & =0.9 \mathrm{~V} \end{aligned} $

Hint

Energy of $n$-photon,

$ E_n=n h v=n h f $

Energy of $2 n$ photons,

$ E_{2 n}=2 n h(3 f)=6 n h f $

$ \frac{\text { Average radiation intensity on surface } A}{\text { Average radiation intensity on surface } B} $

$ \begin{aligned} & =\frac{E_n / t_1}{E_{2 n} / t_2}=\frac{E_n \times t_2}{E_{2 n} \times t_1} \\ & =\frac{(n h f) \times 4 t}{6 n h f \times t}=\frac{4}{6}=\frac{2}{3} \text { or } 2: 3 \end{aligned} $

Energy of photon released by the transition of an electron from second excited state $\left(n_2=3\right)$ of ground state $\left(n_1=1\right)$ is given as

$ \begin{aligned} \Delta E & =h v=E_3-E_1 \\ & =-\frac{13.6}{3^2}-\left(-\frac{13.6}{1^2}\right) \\ & =13.6-\frac{13.6}{9} \\ & =\frac{8}{9} \times 13.6 \mathrm{eV} \end{aligned} $

According to Einstein's photoelectric equation

$ \begin{aligned} \mathrm{KE} & =h v-\phi_0 \\ & =\frac{8}{9} \times 13.6-314 \approx 10 \mathrm{eV} \end{aligned} $

de-Broglie wavelength of the most energetic electron,

$ \begin{aligned} \lambda & =\frac{h}{\sqrt{2 m(K E)}} \\ & =\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.31 \times 10^{-31} \times 10 \times 1.6 \times 10^{-19}}} \\ & =4.09 \times 10^{-10} \mathrm{~m}=4.09 \mathop {\rm{A}}\limits^{\rm{o}}=\simeq 4 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

Given speed of moving particle

$ y=0.8 \mathrm{c} $

We know that, de-Broglie wavelength associated with moving particle

$ \begin{aligned} \lambda & =\frac{h}{m v} \\ \lambda & =\frac{h}{m \times 0.8 c} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

de-Broglie wavelength of photon

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

Energy of photon, $E=\frac{h c}{\lambda}$

Kinetic energy of particle $E^{\prime}=\frac{1}{2} m v^2$

$ \begin{aligned} & =\frac{1}{2} m \cdot(0.8 c)^2=0.32 m c^2 \\ \therefore \quad \frac{E}{E^{\prime}} & =\frac{h c / \lambda}{0.32 m c^2}=\frac{h}{\lambda(0.32 m c)} \\ & =\frac{h}{\frac{h}{m \times 0.8 c} \times 0.32 m c}(\text { from Eq. }(\mathrm{i})) \\ & =\frac{0.8}{0.32}=\frac{5}{2} \text { or }=5: 2 \end{aligned} $

Given:

Initial wavelength: $\lambda_1 = 1 \text{ nm}$

Final wavelength: $\lambda_2 = 0.5 \text{ nm}$

According to de-Broglie's principle, the wavelength ($\lambda$) is given by:

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

where $p$ (momentum) is related to the kinetic energy ($K$) by:

$ p = \sqrt{2mK} $

This leads to the expression for the kinetic energy in terms of wavelength:

$ K = \frac{h^2}{2m\lambda^2} $

where:

$p$ = momentum

$K$ = kinetic energy

$m$ = mass of the electron

$h$ = Planck's constant

Initial Kinetic Energy:

$ K_i = \frac{h^2}{2m(\lambda_1)^2} = \frac{h^2}{2m(1 \text{ nm})^2} $

Final Kinetic Energy:

$ K_f = \frac{h^2}{2m(\lambda_2)^2} = \frac{h^2}{2m(0.5 \text{ nm})^2} $

Change in Kinetic Energy:

The change in kinetic energy $\Delta K$ is calculated as:

$ \Delta K = K_f - K_i = \frac{h^2}{2m} \left(\frac{1}{(0.5)^2} - \frac{1}{1^2}\right) $

Simplifying, we get:

$ \Delta K = 3 \left(\frac{h^2}{2m}\right) = 3K_i $

Thus, the additional energy required to halve the wavelength is three times the initial energy.

The de Broglie wavelength of an electron, which is accelerated by a potential difference $ V $, is calculated using the following formula:

$ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m E_k}} $

where $ E_k = eV $ (with $ e $ being the charge of the electron and $ V $ the potential difference). This leads to:

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

By substituting the known values of $ h $, $ m $, and $ e $ into the formula, we simplify it to:

$ \lambda = \frac{1.23}{\sqrt{V}} \text{ nm} $

To find the de Broglie wavelength for a potential difference of 900 V, we calculate:

$ \lambda = \frac{1.23}{\sqrt{900}} \text{ nm} = \frac{1.23}{30} \text{ nm} $

Thus, the de Broglie wavelength is approximately:

$ \lambda = 0.04 \text{ nm} $

As we know,

$\begin{aligned} F & =q v B=m \frac{v^2}{R} \Rightarrow v=\frac{q}{m} B R . \\ \mathrm{K} . \mathrm{E} . & =\frac{1}{2} m v^2=\frac{1}{2} \frac{e^2 B^2 R^2}{m}=74.4 \mathrm{~keV} . \end{aligned}$

Energy of a photon $=\frac{h c}{\lambda}=49.5 \mathrm{~keV}$.

Binding energy $=74.4-49.5 \mathrm{~keV}=24.9 \mathrm{~keV}$

To determine the velocity of the electron emitted from the gun, we need to apply the concept of energy conservation. Specifically, the kinetic energy (KE) gained by the electron will be equal to the electrostatic potential energy (PE) provided by the potential difference (V) between the filament and the plate.

The electrostatic potential energy can be expressed as:

$ PE = eV $

where:

• $ e $ is the charge of the electron, approximately $ 1.6 \times 10^{-19} \text{ C} $
• $ V $ is the potential difference, which is given as $ 4000 \text{ V} $

The kinetic energy of the electron can be expressed as:

$ KE = \frac{1}{2} mv^2 $

where:

• $ m $ is the mass of the electron, approximately $ 9.1 \times 10^{-31} \text{ kg} $
• $ v $ is the velocity of the electron we need to determine.

Equating the kinetic energy to the potential energy, we have:

$ eV = \frac{1}{2} mv^2 $

Rearranging the equation to solve for the velocity $ v $ gives:

$ v = \sqrt{\frac{2eV}{m}} $

Substituting the given values:

$ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 4000}{9.1 \times 10^{-31}}} $

$ v = \sqrt{\frac{1.28 \times 10^{-15}}{9.1 \times 10^{-31}}} $

$ v = \sqrt{1.41 \times 10^{15}} $

$ v \approx 3.75 \times 10^7 \text{ m/s} $

Therefore, the correct option is:

Option D: $ 3.75 \times 10^7 \mathrm{~m} / \mathrm{s} $

To determine the rate of change of the de-Broglie wavelength of the electron at time $t$, we need to start with the expression for the de-Broglie wavelength, which is given by

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

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

Since the electron experiences a constant electric field of $2E$, it undergoes a uniform acceleration. The force $ F $ on the electron is given by

$F = e \cdot 2E = 2eE$

This force causes the electron to accelerate, so the acceleration $a$ is given by

$a = \frac{F}{m} = \frac{2eE}{m}$

Initially, the electron is at rest, so its velocity $v$ at time $t$ can be obtained using the equation of motion:

$v = u + at = 0 + \left(\frac{2eE}{m}\right) t = \frac{2eE t}{m}$

The momentum $p$ of the electron at time $t$ is given by

$p = mv = m \left( \frac{2eE t}{m} \right) = 2eE t$

Now substituting the expression for momentum into the de-Broglie wavelength equation, we get

$\lambda = \frac{h}{p} = \frac{h}{2eE t}$

To find the rate of change of de-Broglie wavelength with respect to time, we take the derivative of $\lambda$ with respect to $t$:

$\frac{d\lambda}{dt} = \frac{d}{dt} \left( \frac{h}{2eE t} \right) = h \cdot \frac{d}{dt} \left( \frac{1}{2eE t} \right) = h \cdot \left( \frac{-1}{(2eE t)^2} \cdot 2eE \right) = \frac{-h}{2eE} \cdot \frac{1}{t^2} \cdot 2eE = \frac{-h}{t^2}$

Thus, the rate of change of de-Broglie wavelength of the electron at time $t$ is

$\frac{d\lambda}{dt} = \frac{-h}{2eE t^2}$

So the correct answer is:

Option A: $-\frac{h}{\left(2eEt^2\right)}$

We know,

Momentum $(p) = \sqrt {2m{E_k}} $

and ${E_k} = q{V_{acc}}$

$\therefore$ $p = \sqrt {2mq\,{V_{acc}}} $

Both $\alpha$ particle and proton are passed through same potential difference.

$\therefore$ ${\left( {{V_{acc}}} \right)_\alpha } = {\left( {{V_{acc}}} \right)_p} = v$

$\therefore$ ${p_\alpha } = \sqrt {2{m_\alpha }{q_\alpha }v} $

${p_p} = \sqrt {2{m_p}{q_p}v} $

$\therefore$ ${{{p_\alpha }} \over {{p_p}}} = \sqrt {{{{m_\alpha }{q_\alpha }} \over {{m_p}{q_p}}}} $

$ = \sqrt {{{4{m_p} \times 2e} \over {{m_p} \times e}}} $

$ = \sqrt {{8 \over 1}} $

$ = {{2\sqrt 2 } \over 1}$

$k = {{hc} \over \lambda } - \phi = E$

and, $2k = {{hc} \over {{\lambda _2}}} - \phi = 2E$

$ \Rightarrow {{hc} \over \lambda } - E = {{hc} \over {{\lambda _2}}} - 2E$

$ \Rightarrow {{hc} \over {{\lambda _2}}} = {{hc} \over \lambda } + E$

$ \Rightarrow {\lambda _2} = {{hc\lambda } \over {hc + \lambda E}}$

${1 \over 2}mv_1^2 = 5\phi - \phi $

And, ${1 \over 2}mv_2^2 = 10\phi - \phi $

$ \Rightarrow {\left( {{{{v_1}} \over {{v_2}}}} \right)^2} = {4 \over 9}$

$ \Rightarrow {{{v_1}} \over {{v_2}}} = {2 \over 3}$

The de Broglie wavelength of a particle can be expressed as:

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

where:

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

For an electron accelerated through a potential difference of $ V $ volts, its kinetic energy $ K $ is given by:

$ K = eV $

where $ e $ is the charge of an electron.

The electron's momentum can be expressed in terms of its kinetic energy as:

$ p = \sqrt{2mK} $

where $ m $ is the mass of the electron.

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

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

Comparing this with the given equation $ \lambda = \frac{1.227}{x} \, nm $, we see that $ x $ must correspond to:

$ x = \sqrt{2mK} $

Since $ K = eV $, we can substitute this into our expression for $ x $ to get:

$ x = \sqrt{2meV} = \sqrt{V} $

where we've used the fact that the mass and charge of an electron are constants.

$\because$ ${1 \over 2}mv_m^2 = h\nu - \phi $

$ \Rightarrow v_m^2$ varies linearly with frequency.

And, threshold wavelength can be explained by particle nature of light.

$\lambda$ = 900 nm

I = 100 W/m²

A = 10^$-$4

$\Rightarrow$ P = 10^$-$2 W

$\Rightarrow$ Number of photons incident per second

$ = {{{{10}^{ - 2}}\lambda } \over {hc}}$

$ = {{9 \times {{10}^{ - 11}} \times {{10}^2}} \over {6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} \simeq 4.5 \times {10^{16}}$

$\lambda = {h \over {mv}} = {h \over {\sqrt {2m\,eV} }}$

so ${{{\lambda _p}} \over {{\lambda _d}}} = {{\sqrt {{m_d}{V_d}} } \over {\sqrt {{m_p}{V_p}} }} = {1 \over {\sqrt 2 }}$

${{2{V_d}} \over {{V_p}}} = {1 \over 2}$

${{{V_p}} \over {{V_d}}} = {4 \over 1}$

$\because$ ${K_m} = {{hc} \over \lambda } - \phi $

$ \Rightarrow K = {{1230} \over {800}} - \phi $

and, $2K = {{1230} \over {500}} - \phi $

$ \Rightarrow 2 \times {{1230} \over {800}} - 2\phi = {{1230} \over {500}} - \phi $

$ \Rightarrow \phi = 0.615\,eV$

Frequency of EM waves = ${6 \over {2\pi }} \times {10^{15}}$ and ${9 \over {2\pi }} \times {10^{15}}$

Energy of one photon of these waves

$ = \left( {4.14 \times {{10}^{ - 15}} \times {6 \over {2\pi }} \times {{10}^{15}}} \right)$ eV

and $\left( {4.14 \times {{10}^{ - 15}} \times {9 \over {2\pi }} \times {{10}^{15}}} \right)$ eV

= 3.95 eV and 5.93 eV

$\Rightarrow$ Energy of maximum energetic electrons

= 5.93 $-$ 2.50 = 3.43 eV

When energy of incident radiation is equal to the work function of the metal, then the KE of photoelectrons would be zero. But this statement does not comment on the situation when energy is less than the work function.

${K_1} = {{hc} \over {{\lambda _1}}} - \phi = {{hc} \over {3{\lambda _2}}} - \phi $ ..... (i)

and ${K_2} = {{hc} \over {{\lambda _2}}} - \phi $ ..... (ii)

from (i) and (ii) we can say

$3{K_1} = {K_2} - 2\phi $

${K_1} < {{{K_2}} \over 3}$

${\lambda _p} = {h \over p} = {{hc} \over E}$ ...... (i)

${\lambda _e} = {h \over {\sqrt {2mE} }}$ ...... (ii)

From (i) and (ii)

${\lambda _p} \propto \lambda _e^2$

${K_\alpha } = {K_C}$

${{p_\alpha ^2} \over {2{m_\alpha }}} = {{p_C^2} \over {2{m_C}}}$

${{{p_\alpha }} \over {{p_C}}} = \sqrt {{{{m_\alpha }} \over {{m_C}}}} $

So ${{{\lambda _\alpha }} \over {{\lambda _C}}} = {{h/{p_\alpha }} \over {h/{p_C}}} = \sqrt {{{{m_C}} \over {{m_\alpha }}}} $

So ${{{\lambda _\alpha }} \over {{\lambda _C}}} = \sqrt 3 $

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

$R = {{mv} \over {Bq}} = {{\sqrt {2m(KE)} } \over {Bq}}$ ...... (ii)

Putting the values,

$\phi \simeq 1.36$ eV