1.

The work function of Caesium metal is 2.14ev . A beam of light of frequency 6 ×1014 Hz easy incident on metal surface. Find a) energy of incident Photon.b) maximum kinetic energy of photoelectron. Given, h = 6.63 × 10-34 JS

Answer»

Given, \(\phi=2.14\,ev\) 

v = 6 x 1014

h = 6.63 x 10-34 55 Js

(a) Energy of photon E = hv

E = 6.63 x 10-34 x 6 x 1014

E = \(\frac{39.78\times10^{-20}}{1.6\times10^{-19}}\) 

E = 24.86 x 10-1

E = 2.48 ev

(b) Maximum kinetic energy of photo electron

k = hv - \(\phi\)

k = 2.48 - 2.14

k = 0.34 ev



Discussion

No Comment Found

Related InterviewSolutions