Question

If ${\lambda _0}$ and $\lambda$ be the threshold wavelength and the wavelength of the incident light respectively, then the maximum velocity of photo-electrons ejected from the metal surface is:
(A) $\sqrt {\dfrac{{2h}}{m}({\lambda _0} - \lambda )}$
(B) $\sqrt {\dfrac{{2hc}}{m}(\dfrac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}})}$
(C) $\sqrt {\dfrac{{2hc}}{m}({\lambda _0} - \lambda )}$
(D) $\dfrac{{2h}}{m}[\dfrac{1}{{{\lambda _0}}} - \dfrac{1}{\lambda }]$

Answer 1

According to the question, first we should know about the formation of photo-electrons and the minimum and maximum energy forms by the electrons respectively. And then find the maximum velocity of that photo-electrons during ejection from the metal surface.

Complete step-by-step answer: When light is inclined on a metal surface, it forms electrons and after ejection, we can call the electrons as Photo-electrons, and this effect is known as photo-electrons effect. And the inclined electrons kept minimum energy and after achieving minimum energy, the metal surface can only eject photo-electron, & that energy is known as Work-function, $w$ :
$\because w = h{v_0}$
here, ${v_0}$ is threshold frequency;
Threshold frequency is defined as the minimum frequency of incident light which can cause photoelectric emission i.e. this frequency is just able to eject electrons without giving them additional energy.
$\Rightarrow w = \dfrac{{hc}}{{{\lambda _0}}}$
Here, ${\lambda _0}$ is the threshold wavelength.
Now,
When light is inclining on the metal surface, it had its energy, $e$ :
$\because e = hv = \dfrac{{hc}}{\lambda }$
here, $\lambda$ is the actual wavelength .
After inclination, the strength of energy decreases. So, $e$ is more than $w$ .
$\therefore$ Kinetic Energy = Energy-Work Function
$\because K.E = e - w$
$\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}}$
here, m=mass and v= maximum velocity
$\Rightarrow v = \sqrt {\dfrac{{2hc}}{m}(\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}})}$
Finally,
$\therefore v = \sqrt {\dfrac{{2hc}}{m}(\dfrac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}})}$

Hence, the correct option is B. $\sqrt {\dfrac{{2hc}}{m}(\dfrac{{{\lambda _0} - \lambda }}{{\lambda {\lambda _0}}})}$ .

Note: The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 vol. When the radiation 256.7 nm is used.