Question

If the energy of incident photon and work function of metal are E eV and ${\varphi }_{0}eV$ respectively, then the maximum velocity of emitted photoelectron will be

• A. $\frac{2}{m}[E-{\varphi }_0]$
• B. $\sqrt{\frac{2}{m}(E-{\varphi }_0)}$
• C. $\frac{m}{2}[E-{\varphi }_0]$
• D. $2m\sqrt{(E-{\varphi }_0)}$

Answer 1

The correct option is B $\sqrt{\frac{2}{m}(E-{\varphi }_0)}$

The maximum energy of emitted photoelectron is
$E_{max}=h\nu -\varphi$
where, $h\nu$ is the energy of incident photon, $\varphi$ is the photoelectric work function of the metal.
For an electron having mass m and maximum velocity v we can write

$\frac{1}{2}m{v}_{max}^2=E-{\varphi }_0$ .................(given, $h\nu =E$)

$v_{max}^2=\frac{2}{m}(E-{\varphi }_0)$

$v_{max}=\sqrt{\frac{2}{m}(E-{\varphi }_0)}$

So, the answer is option (B).