Dual Nature of Radiation

As $\lambda $ decreases, $v$ increases and hence the speed of photo electron increases. The chances of photo electron to meet the anode increases and hence photo electric current increases.

Emission of photo-electron starts from the surface after incidence of photons in about ${10^{ - 10}}s.$

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

${I_2} \to 4\,\,$ 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 ${\lambda \over {\sqrt 2 }}$

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 $\left. {\,\,} \right]$

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