Question

Light of wavelength 4000 angstrom is incident on a metal plate whose function is 2 ev . the maximum kinetic energy of emitted photoelectron will be

• A. 12.01 eV
• B. 21.22 eV
• C. 23.12 eV
• D. 10.37 eV

Answer 1

The correct option is D 10.37 eV

We know that,

Maximum energy of emitted photon electron

K.E energy of the incident radiation –work function

Energy of the incident radiation =$\frac{hc}{\lambda }$

Now,

$=\frac{hc}{\lambda }$

$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4000\times {10}^{-10}\times 1.6\times {10}^{-19}}$

$=12.37eV$

Now, the kinetic energy is

$K.E=12.37-2$

$K.E=10.37eV$

Hence, the kinetic energy is $10.37eV$