Question

A proton has kinetic energy $E=100keV$ which is equal to that of a proton. The wavelength of photon is ${\lambda }_2$ and that the proton is ${\lambda }_1$. The ratio of ${\lambda }_2/{\lambda }_1$ is proportional to

• A. $E^2$
• B. $E^{-1/2}$
• C. $E^{-1}$
• D. $E^{1/2}$

Answer 1

The correct option is B $E^{-1/2}$

Here, kinetic energy of proton $=$ energy of photon

So, $E=\frac{hc}{{\lambda }_2}$

or ${\lambda }_2=hc/E$

If $p$ be the momentum of proton, then $E=\frac{p^2}{2m}$ or $p=\sqrt{2mE}$

From De Broglie hypothesis, wavelength of proton is ${\lambda }_1=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

Thus, $\frac{{\lambda }_2}{{\lambda }_1}=\frac{hc}{E}\times \frac{\sqrt{2mE}}{h}=c\sqrt{2m}{E}^{-1/2}\propto E^{-1/2}$