Question

When photons of energy $hc/\lambda$ fall on a metal surface, photoelectrons are ejected from it. If the work function of the surface is $h{v}_0$, then

• A. Maximum kinetic energy of the electron is $\left[\right(hc/\lambda )-h{v}_0]$
• B. Maximum kinetic energy of the photoelectron is equal to $\left(hc/\lambda \right)$
• C. Minimum KE of the photoelectron is zero
• D. Minimum kinetic energy of the photoelectron is equal to $hc/\lambda$

Answer 1

Answer: (A), (C)

The correct options are
A Maximum kinetic energy of the electron is $\left[\right(hc/\lambda )-h{v}_0]$
C Minimum KE of the photoelectron is zero

The Eisenstein's equation for photoelectric effect is $K{E}_{max}=E-W$

Where $K{E}_{max}=$ maximum kinetic energy of photoelectron , $E=$ energy of photon and $W=$ work function of metal.

Given, $E=hc/\lambda$ and $W=h{\nu }_0$

Thus, $K{E}_{max}=hc/\lambda -h{\nu }_0$

For minimum kinetic energy, velocity will be zero . Since $KE=\left(1/2\right)m{v}^2$ so minimum kinetic energy will be zero.