Question

A photon has kinetic energy E = 100 keV which is equal to that of a proton. The wavelength of photon is ${\lambda }_2$ and that of 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 $E^{-1/2}$

For photon: E = hv
or $E=\frac{hc}{\lambda }$
$\Rightarrow {\lambda }_2=\frac{hc}{E}$ (i)
For proton: $E=\frac{1}{2}{m}_p{v}_p^2$
$E=\frac{1}{2}\frac{m_p^2{v}_p^2}{m}\Rightarrow P=\sqrt{2mE}$
From de Broglie equation,
$P=\frac{h}{{\lambda }_1}$
$\Rightarrow {\lambda }_1=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$ (ii)
$\frac{h_2}{{\lambda }_1}=\frac{hc}{E\times \frac{h}{2mE}}\alpha E^{-1/2}$