Question

In a photo-emissive cell with exciting wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to 3$\lambda$/4, the speed of fastest emitted electron will be:
A) $v{(3/4)^{1/2}}$
B) $v{(4/3)^{1/2}}$
C) less than $v{(4/3)^{1/2}}$
D) greater that $v{(4/3)^{1/2}}$

Answer 1

Wavelength decreases if it is changed to 3$\lambda$/4. This means that frequency increases. Therefore, the work function and energy of electrons will also increase. The kinetic energy and velocity of electrons depends on the frequency of incident light. Thus if frequency is increased, velocity should also increase.

Formula used:
Einstein’s photoelectric equation is $hf = {\omega _0} + \dfrac{1}{2}mv_{}^2$

Complete step by step answer:
The ejection of electrons from a photo-emissive cell on absorption of light due to an exciting wavelength is the phenomenon of photoelectric effect.
Suppose, the frequency of incident photons is $f$ and the work function of photoelectric metal is ${\omega _0}$. It is given that the maximum velocity of the ejected electron is $v$ and let the mass of that electron be m. Then, according to Einstein’s photoelectric equation
$hf = {\omega _0} + \dfrac{1}{2}mv_{}^2$ $\to$(1)
where $h$ is the Planck's constant
$\dfrac{1}{2}mv_{}^2$ is the Kinetic energy of photoelectrons.
The work function is the minimum energy required to just eject the photoelectron from the photosensitive material. Therefore, it can be written as
$$$$$${\omega 0} = h{f_0}$$ \to $(2) where${f_0}$is the threshold frequency. Substituting equation (2) in equation(1)$hf = h{f_0} + \dfrac{1}{2}mv{}^2$$ \to $(3) Rearranging the terms in order to isolate ‘$v$’$v = \sqrt {\dfrac{2}{m}h(f - {f_0})} $$ \to $(4) It is given that the exciting wavelength is changed to 3$\lambda $/4. So, let the new wavelength be$\lambda ' = \dfrac{{3\lambda }}{4}$. Also, let the new frequency be$f' = \dfrac{4}{{3\lambda }}$, since frequency is inversely proportional to wavelength. So, the new velocity$v'$will be$v' = \sqrt {\dfrac{2}{m}h(\dfrac{1}{{\lambda '}} - {f_0})} = \sqrt {\dfrac{2}{m}h(\dfrac{4}{{3\lambda }} - {f_0})} $$ \to $(5) Dividing equation (5) by (4)$\dfrac{{v'}}{v}=\dfrac{{\sqrt{\dfrac{2}{m}h(\dfrac{4}{{3\lambda}}-{f_0})}}}{{\sqrt{\dfrac{2}{m}h(\dfrac{1}{\lambda } - {f_0})} }} = \sqrt {\dfrac{{\dfrac{4}{{3\lambda }} - {f_0}}}{{\dfrac{1}{\lambda } - {f_0}}}} $$\because {f_0} > 0$and the total frequency is greater than threshold frequency, RHS will be positive.$\therefore v' > \sqrt {\dfrac{4}{3}v} $$

So, the correct answer is “Option D”.

Note:
Photoelectrons are always ejected with velocities that are in a range from 0 to $v$ . No photoelectric emission occurs if the frequency of incident radiation is lower than threshold frequency. Velocity of photoelectrons depends on frequency of incident light and not on the intensity of incident light.