Dual Nature of Radiation

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 .

$I \propto {1 \over {{r^2}}};{{{I_1}} \over {{I_2}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^2} = {1 \over 4}$

${I_2} \to 4\,\,$ times ${I_1}$

When intensity becomes 4 times, no. of photoelectrons emitted would increase by $4$ times, since number of electrons emitted per second is directly proportional to intensity.

de-Broglie wavelength,

$\lambda = {h \over p} = {h \over {\sqrt {2.m,\left( {K.E} \right)} }}$

$\therefore$ $\lambda \propto {1 \over {\sqrt {K.E} }}$

If $K.E$ is doubled, wavelength becomes ${\lambda \over {\sqrt 2 }}$

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

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

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 $

For the longest wavelength to emit photo electron

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

$ \Rightarrow \lambda = {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {40 \times 1.6 \times {{10}^{ - 16}}}} = 310nm$

Einstein's photoelectric equation is given by

$K_{\max} = h\nu - \phi,$

where $K_{\max}$ is the maximum kinetic energy of the emitted photoelectrons, $h$ is Planck's constant, $\nu$ is the frequency of the incident radiation, and $\phi$ is the work function of the metal (the minimum energy needed to eject an electron).

If you plot $K_{\max}$ against $\nu$, you will get a straight line with slope $h$ (Planck's constant) and y-intercept $-\phi$ (the negative of the work function). The slope of the line (which is $h$) does not depend on the intensity of the radiation or the type of metal used. Instead, it is a universal constant.

Therefore, the correct answer is:

Option D: The slope is the same for all metals and independent of the intensity of the radiation.

For one photo cathode

$h{f_1} - W = {1 \over 2}mv_1^2\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

For another photo cathode

$h{f_2} - W = {1 \over 2}mv_2^2\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

Subtracting $(ii)$ from $(i)$ we get

$\left( {h{f_1} - W} \right) - \left( {h{f_2} - W} \right) = {1 \over 2}mv_1^2 - {1 \over 2}mv_2^2$

$\therefore$ $h\left( {{f_1} - {f_2}} \right) = {m \over 2}\left( {v_1^2 - v_2^2} \right)$

$\therefore$ $v_1^2 - v_2^2 = {{2h} \over m}\left( {{f_1} - {f_2}} \right)$

Formation of covalent bond is best explained by molecular orbital theory.

We know that work function is the energy required and energy $E = h\upsilon $

$\therefore$ ${{{E_{NA}}} \over {{E_{Cu}}}} = {{h{\upsilon _{Na}}} \over {h{\upsilon _{cu}}}} = {{{\lambda _{cu}}} \over {{\lambda _{Na}}}}$

$\left[ {\,\,} \right.$ as ${\,\,\,\upsilon \propto {1 \over \lambda }}$ for light $\left. {\,\,} \right]$

$\therefore$ ${{{\lambda _{Na}}} \over {{\lambda _{Cu}}}} = {{{E_{Cu}}} \over {{E_{Na}}}} = {{4.5} \over {2.3}} \approx {2 \over 1}$