The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_1$ is the de-Broglie…
The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_1$ is the de-Broglie wavelength of a proton, $\lambda_2$ the wavelength associated with the proton and if the energy of the photon is $E$, then $\left(\lambda_1 / \lambda_2\right)$ is proportional to
$E^4$
$E^{1 / 2}$
$E^2$
$E$
Solution
We know that, de-Broglie wavelength of photon
$
E=h \mathrm{v}
$
But we know that also
$
\begin{aligned}
E & =\frac{1}{2} m v^2 \\
\therefore \quad 2 E & =\frac{P^2}{m} \\
\because \quad P & =m v \\
\frac{P^2}{m} & =m v^2 \\
P & =\sqrt{2 m E}
\end{aligned}
$
(for proton)
According to the question
$
\lambda_1=\frac{h}{P}
$
(de-Broglie of matter waves)
and
$
\lambda_2=\frac{h c}{E}
$
From the Eqs. (ii) and (iii), we get
$
\begin{aligned}
\frac{\lambda_1}{\lambda_2} & =\frac{h / P}{h c / E}=\frac{E}{P_C} \\
& =\frac{E}{\sqrt{2 m E} \cdot C}=\frac{\sqrt{E}}{\sqrt{2 m} \cdot C}
\end{aligned}
$
$\begin{array}{ll}\text { So, } & \frac{\lambda_1}{\lambda_2}=\frac{\sqrt{E}}{\sqrt{2 m} \cdot c} \\ & \frac{\lambda_1}{\lambda_2} \propto \sqrt{E} \\ \Rightarrow & \lambda_1: \lambda_2=E^{1 / 2}\end{array}$