Q. 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:
NTA AbhyasNTA Abhyas 2022
Solution:
Given : work function of metal, $\phi=2.28eV$
Wavelength of light, $\lambda =500nm=500\times 10^{- 9}m$
According to Einstein's Photoelectric equation, we have
$K_{m a x}=\frac{h c}{\lambda }-\phi$
$\Rightarrow K_{m a x}=\frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 9} \times 1.6 \times 10^{- 19}}-2.28$
$\Rightarrow K_{m a x}=2.48-2.28=0.2eV$
As de Broglie wavelength is given by $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2 m K}}$
So, de Broglie wavelength of electron is minimum when kinetic energy is maximum.
$\therefore \lambda _{m i n}=\frac{h}{\sqrt{2 m K_{m a x}}}$
$\Rightarrow \lambda _{m i n}=\frac{6 . 6 \times 10^{- 34}}{\sqrt{2 \times 9 \times 10^{- 31} \times 0.2 \times 1.6 \times 10^{- 19}}}$
$\Rightarrow \lambda _{m i n}=2.8\times 10^{- 9}m$
$\therefore \lambda \geq 2.8\times 10^{- 9}m$
Wavelength of light, $\lambda =500nm=500\times 10^{- 9}m$
According to Einstein's Photoelectric equation, we have
$K_{m a x}=\frac{h c}{\lambda }-\phi$
$\Rightarrow K_{m a x}=\frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 9} \times 1.6 \times 10^{- 19}}-2.28$
$\Rightarrow K_{m a x}=2.48-2.28=0.2eV$
So, de Broglie wavelength of electron is minimum when kinetic energy is maximum.
$\therefore \lambda _{m i n}=\frac{h}{\sqrt{2 m K_{m a x}}}$
$\Rightarrow \lambda _{m i n}=\frac{6 . 6 \times 10^{- 34}}{\sqrt{2 \times 9 \times 10^{- 31} \times 0.2 \times 1.6 \times 10^{- 19}}}$
$\Rightarrow \lambda _{m i n}=2.8\times 10^{- 9}m$
$\therefore \lambda \geq 2.8\times 10^{- 9}m$