Question

The work function of a photoelectric material is 3.3 eV. It threshold frequency will be:
A) $4 \times {10^{23}}Hz$
B) $8 \times {10^{12}}Hz$
C) $4 \times {10^{11}}Hz$
D) $8 \times {10^{14}}Hz$

Answer 1

Hint
Threshold frequency is the minimum amount of the frequency of the radiation which is required to produce photoelectric effect. It is the ratio of the work function and Planck’s constant. As the value of the work function is given in eV first convert it into joule and then divide the Planck’s constant.

Complete Step by step solution
It is given that work function $\phi = 3.3eV$
First of all, convert the eV into joule by multiplying it with , we get
$ \Rightarrow \phi = 3.3 \times 1.6 \times {10^{ - 19}}J$ …………………….. (1)
As threshold frequency is the minimum frequency of radiation which is required for the photoelectric effect. It is the ratio of work function and Planck’s constant.
$ \Rightarrow {\nu _0} = \dfrac{\phi }{h}$ ……………………….. (2)
Where, ϕ is the work function
${\nu _0}$is the threshold frequency
h is the Planck’s constant. Its value is $h = 6.634 \times {10^{ - 34}}Js$
put the value of equation (1) and Planck’s constant in equation (2), we get
$ \Rightarrow {\nu _0} = \dfrac{{3.3 \times 1.6 \times {{10}^{ - 19}}}}{{6.634 \times {{10}^{ - 34}}}}$
$ \Rightarrow {\nu _0} = 8 \times {10^{14}}Hz$
Hence this is the required threshold frequency for the photoelectric effect.
Thus, option (D) is correct.

Note
Here, it should be noticed that the work function is given in electron-volt but as Planck’s constant has unit joule-sec, therefore we must first the work function from electron-volt to Joule after that we solve the problem otherwise, we will get the wrong results.