Question

Light of wavelength $4000\mathring{A}$ is incident on a metal plate whose work function is $2eV$. What is the maximum kinetic energy of emitted photoelectron?

• A. $0.5eV$
• B. $1.1eV$
• C. $2.0eV$
• D. $1.5eV$

Answer 1

Answer: (B)

$1.1eV$

Given: work fuction = $2eV$, $\lambda =4000A^o$

Applying Einstein's equation,

$hv=\varphi +\frac{1}{2}m{v}^2=\varphi +K.E=\frac{hc}{\lambda }+K.E.$

$6.6\times {10}^{-34}\times \frac{3\times {10}^8}{4000\times {10}^{10}}+K.E.$

$=2\times 1.6\times {10}^{-19}+K.E$

$\frac{6.6\times 3}{4}\times {10}^{19}=3.2\times {10}^{-19}+K.E$
$4.95\times {10}^{-19}=3.2\times {10}^{-19}+K.E$

$K.E=(4.95-3.2)\times {10}^{-19}$

$=1.75\times {10}^{-19}J$.

$=\frac{1.75\times {10}^{-19}}{1.6\times {10}^{-19}}eV=1.1eV$