Question

An electron and a photon, each has a de-Broglie wavelength of $1.2A$. The ratio of their energies will be

• A. $1:1$
• B. $1:10$
• C. $1:100$
• D. $1:1000$

Answer 1

The correct option is C $1:100$

For photon the deBroglie wavelength is $\lambda =\frac{h}{p}=\frac{h}{E_{ph}/c}=\frac{hc}{E_{ph}}$ ...(1_)

and for an electron the de-broglie wavelength is given as $\lambda =\frac{h}{p}=\frac{h}{\sqrt[2]{22m{E}_e}}$

on squaring we get ${\lambda }^2=\frac{h^2}{2m{E}_e}$ .............(2)

dividing equation-1 by equation-2 , we get $\frac{1}{\lambda }=2mc\frac{E_e}{h{E}_{ph}}$

putiing $\lambda =1.2\times {10}^{-10}meter$, $m=9.1\times {10}^{-31}Kg$ , $c=3\times {10}^{8}m/s$ as speed of light

and $h=6.62\times {10}^{-34}Js$

we get $\frac{E_e}{E_{ph}=\frac{1}{100}}$