Light of wavelength 4000a is incident on a metal plate whose work function is 2 ev what is maximum kinetic energy of emitted photoelectron
Answers
Answered by
3
Answer:
The maximum kinetic energy is 10.37 eV
Explanation:
According to the problem the wavelength of the light i 4000a
The work function of the plate is 2 ev
The maximum kinetic energy emitted,
KE=Energy of the incident radiation −work functionEnergy of the incident radiation
=hc/λ [ where h = plank's constant, c= speed of light, λ= wavelength]
=6.6×10^(−34)×3×10^8/4000×10^(−10) joule
=6.6×10^(−34)×3×10^8/4000×10^(−10)×1.6×10^(−19)
=12.37eV
KE=12.37−2=10.37 eV
Similar questions