Question

1. What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.8×1015Hz?

Express your answer in joules to three significant figures.

2. A green laser pointer emits light with a wavelength of 521 nm. What is the frequency of this light

Answer 1

(1)

Threshold frequency of electron in Cs = 9.39 * 1014 Hz

Frequency of photon used = 1.8 * 1015 Hz

According to Einstein's photo electric effect,

h$\vartheta$ = h$\vartheta$0 + K.E

K.E = (6.626 * 10^-34 * 1.8 * 10¹⁵) - (6.626*10^-34 * 9.39 * 10¹⁴)

K.E = (11.93 * 10^-19) - (62.22 * 10^-20)

K.E = 5.7 * 10^-19 J

(2) $\vartheta$ = c / Lambda

$\vartheta$ = (3 * 10⁸) m/s / (521 * 10^-9) m

$\vartheta$ = 5.76 * 10¹⁴ Hz