Get Answers to all your Questions

header-bg qa

Q: 11.8 The threshold frequency for a certain metal is  3.3\times 10^1^4\hspace{1mm}Hz  . If light of frequency   8.2\times 10^1^4\hspace{1mm}Hz  is incident on the metal, predict the cutoff voltage for the photoelectric emission. 

Answers (1)

best_answer

Threshold frequency of the given metal(\nu _{0})=  3.3\times 10^1^4\hspace{1mm}Hz

The work function of the given metal is

\\\phi _{0}=h\nu _{0}\\ \phi _{0}=6.62\times 10^{-34}\times 3.3\times 10^{-14}\\ \phi _{0}=2.18\times 10^{-19}\ J

The energy of the incident photons

\\E=h\nu \\ E=6.62\times 10^{-34}\times 8.2\times 10^{14}\\ E=5.42\times 10^{-19}\ J

Maximum Kinetic Energy of the ejected photoelectrons is

\\E-\phi _{0}=3.24\times 10^{-19}\ J\\ E-\phi _{0}=2.025\ eV

Therefore the cut off voltage is 2.025 eV

Posted by

Sayak

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads