1.

Work function of a metal surface is `phi=1.5eV`. If a light of wavelength 5000Å falls on it then the maximum K.E. of ejected electron will be-A. 1.2 eVB. 0.98 eVC. 0.45 eVD. 0 eV

Answer» Correct Answer - B
`K.E_(max)=(hc)/lamda-phi`
=(12400eVÅ)/(5000Å)-1.5eV
=(2.48-1.5)eV=0.98 eV


Discussion

No Comment Found

Related InterviewSolutions