Question

Assuming photo-emission to take place, the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times, is (assuming work function to be negligible in comparison to \[\dfrac{{hc}}{\lambda }\])
A. \[4\]
B. \[\dfrac{1}{4} \\ \]
C. \[2\]
D. \[\dfrac{1}{2}\]

Answer 1

Hint:The equation for photoelectric effect is \[h\nu = {\phi _o} + {E_k}\]. The minimum amount of energy required for electron emission from the metal surface is the work function of that metal and calculated as \[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _0}}}\]. \[{E_k}\] is the kinetic energy of the emitted photoelectron such that \[{E_k} = \dfrac{1}{2}m{v^2}\] where v is the velocity of the electron. If we neglect work function, then on comparing the formulae for energy of photon and the kinetic energy of electron, the velocity of the photoelectron and the wavelength of the incident light are found to be inversely related.

Formula(e) used:
Energy, \[E = h\nu = \dfrac{{hc}}{\lambda }\]
Equation for photoelectric effect, \[h\nu = {\phi _o} + {E_k}\]
Work function, \[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _o}}}\]
kinetic energy of the emitted photoelectron,
\[{E_k} = \dfrac{1}{2}m{v^2}\]
Where,
E = energy of incident radiation
\[{\phi _o}\]= Work function of the metal
\[{E_k}\]= energy of the emitted photoelectron
\[c{\rm{ }} = \text{speed of light}= 3 \times {10^8}m/s\]
\[\nu \]= frequency of the light,
\[{\nu _o}\]= threshold frequency,
\[\lambda \]= wavelength of the light,
\[{\lambda _o}\]= threshold wavelength
\[e = 1.6 \times {10^{ - 19}}C\]= electronic charge
m = mass of the photoelectron
\[v\]= velocity of the photoelectron

Complete step by step solution:
Given: In this photoelectric experiment, wavelength of light is \[\lambda \] increased by a factor of 4 and \[{\phi _o} < < \dfrac{{hc}}{\lambda }\]. So, it can be neglected. We need to determine the change in velocity of this metal.
Equation for photoelectric effect is
\[h\nu = {\phi _o} + {E_k}\]
\[\Rightarrow \dfrac{{hc}}{\lambda } = {\phi _o} + {E_k} \\ \]---- (1)
\[\Rightarrow {\phi _o} < < \dfrac{{hc}}{\lambda } \\ \]---(2)
So work functions can be neglected.
\[{E_k} = \dfrac{1}{2}m{v^2}\]----(3)
From equations (1) ,(2) and (3),
\[\dfrac{{hc}}{\lambda } = {E_k} = \dfrac{1}{2}m{v^2}\]---(4)

If wavelength is increased by a factor of 4. Then, the new incident energy,
\[E' = \dfrac{{hc}}{{4\lambda }} = \dfrac{E}{4} \\ \]----(5)
From equations (4) and (5),
The new kinetic energy is,
\[{E_k}' = \dfrac{{{E_k}}}{4} \\ \]
\[\Rightarrow \dfrac{1}{2}m{v^{'2}} = \dfrac{{\dfrac{1}{2}m{v^2}}}{4} \\ \]
\[\therefore v' = \dfrac{v}{2}\]
So, the maximum velocity will become half the original velocity if the wavelength of the incident radiation is increased by 4 times.

Hence option D is the correct answer.

Note: A direct correlation between velocity and wavelength can only be obtained if these two are sufficiently higher than the work function or the threshold is significantly of lesser value. The remaining energy from the incidence energy after the work function is converted to maximum kinetic energy to enable electron emission from the metal surface. The velocity related to that is the velocity of the photoelectron. This phenomenon is used in devices such as photocell to use incident radiation to generate current.