Physics - Modern Physics Question with Solution | TestHub

Let the kinetic energy of the proton be $K$. We are given $K=E$.

The de-Broglie wavelength of the proton is given by:
${\lambda }_1=\frac{h}{p}=\frac{h}{\sqrt{2mK}}=\frac{h}{\sqrt{2mE}}$

The energy of the photon is given by:
$E=\frac{hc}{{\lambda }_2}\Rightarrow {\lambda }_2=\frac{hc}{E}$

Taking the ratio of ${\lambda }_1$ to ${\lambda }_2$:
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}}=\frac{h}{\sqrt{2mE}}\times \frac{E}{hc}=\frac{1}{c\sqrt{2m}}{E}^{1/2}$

Since $c$ and $m$ are constants, we have:
$\frac{{\lambda }_1}{{\lambda }_2}\propto E^{1/2}$

Comparing this with $\frac{{\lambda }_1}{{\lambda }_2}\propto E^n$, we get $n=0.5$.

Answer: 0.5