Question

If the energy of incident photon and work function of a metal are $E$, ${\varphi }_0$ $eV$ respectively, then the maximum velocity of emitted photoelectrons 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

Answer: (B)

$\sqrt{\frac{2}{m}[E-{\varphi }_0]}$

$h\nu =h{\nu }_0+K.E$
$\Rightarrow E=\varphi +K.E$
$K.E=E-\varphi$
$\frac{1}{2}m{v}^2=E-\varphi$
$v^2=\frac{2}{m}(E-\varphi )$
$v=\sqrt{\frac{2}{m}(E-\varphi )}$