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 $\sqrt 3 v$.

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.

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.