Question

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

• A. $-\ge 2.8\times 10^{-9}m$
• B. $\le 2.8\times 10^{-12}m$
• 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
$K_{\max }=\frac{hc}{\lambda }-{\varphi }_0$
where $\lambda$ is the wavelength of incident light and ${\varphi }_0$ is the work function.
Here, $\lambda =500nm$, he $=1240eVnm$
and ${\varphi }_0=2.28eV$
$\therefore K_{\max }=\frac{1240eVnm}{500nm}-2.28eV$
$=2.48eV-2.28eV=0.2eV$
The de Broglie wavelength of the emitted electron is
${\lambda }_{\min }=\frac{h}{\sqrt{2m{K}_{\max }}}$
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 $K_{\max }=0.2eV=0.2\times 1.6\times 10^{-19}J$
$\therefore {\lambda }_{\min }=\frac{6.6\times 10^{-34}Js}{\sqrt{2\left(9\times 10^{-31}kg\right)\left(0.2\times 1.6\times 10^{-19}J\right)}}$
$=\frac{6.6}{2.4}\times 10^{-9}m=2.8\times 10^{-9}m$
So, $\lambda \ge 2.8\times 10^{-9}m$