Physics - Modern Physics Question with Solution | TestHub

For a photon, the energy is given by $E_p=\frac{hc}{{\lambda }_p}$.

For an electron, the kinetic energy is $E_e=\frac{1}{2}m{v}^2$ and its de-Broglie wavelength is ${\lambda }_e=\frac{h}{mv}$.

Given that the de-Broglie wavelength of the photon is twice that of the electron:
${\lambda }_p=2{\lambda }_e$

Substituting ${\lambda }_e$ into the equation for $E_p$:
$E_p=\frac{hc}{2{\lambda }_e}=\frac{hc}{2\left(\frac{h}{mv}\right)}=\frac{mvc}{2}$

Now, taking the ratio of the energy of the photon to the kinetic energy of the electron:
$\frac{E_p}{E_e}=\frac{\frac{mvc}{2}}{\frac{1}{2}m{v}^2}=\frac{c}{v}$

We are given that the speed of the electron is $v=\frac{c}{100}$.

Therefore, the ratio is:
$\frac{E_p}{E_e}=\frac{c}{\frac{c}{100}}=100=10^2$

Answer: $10^2$