Question

Light of wavelength $500 \,nm$ is incident on a metal with work function $2.28 \,eV$. The de Broglie wavelength of the emitted electron is

• A. $-\ge2.8 \times 10^{-9}m $
• B. $\le2.8 \times 10^{-12}m $
• C. $<2.8 \times 10^{-10}m $
• D. $<2.8 \times 10^{-9}m $

Answer 1

Answer: (A)

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted electron is
$K_{\max }=\frac{h c}{\lambda}-\phi_{0}$
where $\lambda$ is the wavelength of incident light and $\phi_{0}$ is the work function.
Here, $\lambda=500 \,nm$, he $=1240\, eV \,nm$
and $\phi_{0}=2.28 \,e V$
$\therefore K_{\max }=\frac{1240 eVnm }{500 nm }-2.28\, eV$
$=2.48 eV -2.28 \,eV =0.2 \,eV$
The de Broglie wavelength of the emitted electron is
$\lambda_{\min }=\frac{h}{\sqrt{2 m K_{\max }}}$
where $h$ is the Planck's constant and $m$ is the mass of the electron.
As $h =6.6 \times 10^{-34} Js , m=9 \times 10^{-31} kg$
and $K_{\max }=0.2\, eV =0.2 \times 1.6 \times 10^{-19} J$
$\therefore \lambda_{\min }=\frac{6.6 \times 10^{-34} Js }{\sqrt{2\left(9 \times 10^{-31} kg \right)\left(0.2 \times 1.6 \times 10^{-19} J \right)}}$
$=\frac{6.6}{2.4} \times 10^{-9} m =2.8 \times 10^{-9} m$
So, $\lambda \geq 2.8 \times 10^{-9} \,m$