Physics - Modern Physics Question with Solution | TestHub

For a photon of energy $E$, the wavelength is given by ${\lambda }_{photon}=\frac{hc}{E}$.

For an electron of kinetic energy $E$, the momentum is $p=\sqrt{2mE}$. The de Broglie wavelength of the electron is ${\lambda }_{electron}=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$.

Taking the ratio of the wavelength of the electron to that of the photon, we get:

$\frac{{\lambda }_{electron}}{{\lambda }_{photon}}=\frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}}=\frac{h}{\sqrt{2mE}}\times \frac{E}{hc}=\frac{\sqrt{E}}{c\sqrt{2m}}$

Since $h$, $c$, and $m$ are constants, the ratio is directly proportional to $\sqrt{E}$.

$\frac{{\lambda }_{electron}}{{\lambda }_{photon}}\propto \sqrt{E}$

Thus, the ratio is proportional to $\sqrt{E}:1$.

Answer: $\sqrt{E}:1$