Answer to Question #227209 in Atomic and Nuclear Physics for BENSON

Question

A radiation of frequency 1016Hz falls on a photocathode and ejects electrons with maximum energy of 4.2 X 10-19 J. If the frequency of radiation is changed to 5 X 1014 Hz, the maximum energy of ejected electrons becomes 0.9 X10-19J.

The planks’ constant.

(ii) The threshold frequency.

(iii) The work function.

Answer 1

Given:

$f_1=10^{15}\:{\rm Hz},\quad E_{k1}=4.2*10^{-19}\:\rm J$

$f_2=5*10^{14}\:{\rm Hz},\quad E_{k2}=0.9*10^{-19}\:\rm J$

The Einstein's equation for photoelectric effect says

$$hf=\phi+E_k$$

We get:

(i) the Planks’ constant

$$h=\frac{E_{k1}-E_{k2}}{f_1-f_2}\\ =\frac{4.2*10^{-19}\:\rm J-0.9*10^{-19}\:\rm J}{10^{15}\:{\rm Hz}-5*10^{14}\:{\rm Hz}}=6.6*10^{-34}\:\rm J/s$$

(ii) the threshold frequency

$$f_{\rm th}=\frac{\phi}{h}=\frac{hf_1-E_{k1}}{h}\\=\frac{6.6*10^{-34}\:\rm J/s*10^{15}\:{\rm Hz}-4.2*10^{-19}\:\rm J}{6.6*10^{-34}\:\rm J/s}=3.6*10^{14}\:\rm Hz$$

(iii) the work function

$$\phi=hf_{\rm th}=3.6*10^{14}\:\rm Hz*6.6*10^{-34}\:\rm J/s=2.4*10^{-19}\:\rm J$$