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. $<2.8\times 10^{-9}m$
• B. $\ge 2.8\times 10^{-9}m$
• C. $\le 2.8\times 10^{-12}m$
• D. $<2.8\times 10^{-10}m$

Answer 1

Answer: (B)

Given : work function of metal, $\varphi =2.28eV$
Wavelength of light, $\lambda =500nm=500\times 10^{-9}m$
According to Einstein's Photoelectric equation, we have
$K_{max}=\frac{hc}{\lambda }-\varphi$
$\Rightarrow K_{max}=\frac{6.6\times 10^{-34}\times 3\times 10^8}{500\times 10^{-9}\times 1.6\times 10^{-19}}-2.28$
$\Rightarrow K_{max}=2.48-2.28=0.2eV$

As de Broglie wavelength is given by $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$
So, de Broglie wavelength of electron is minimum when kinetic energy is maximum.
$\therefore {\lambda }_{min}=\frac{h}{\sqrt{2m{K}_{max}}}$
$\Rightarrow {\lambda }_{min}=\frac{6.6\times 10^{-34}}{\sqrt{2\times 9\times 10^{-31}\times 0.2\times 1.6\times 10^{-19}}}$
$\Rightarrow {\lambda }_{min}=2.8\times 10^{-9}m$
$\therefore \lambda \ge 2.8\times 10^{-9}m$