Question 291: Dual Nature of Matter and Radiation

Question

Kinetic energy of a proton is equal to energy $E$ of a photon. Let ' $\lambda_1$ ' be the de-Broglie wavelength of proton and ' $\lambda_2$ ' be the wavelength of photon. If $\left(\frac{\lambda_1}{\lambda_2}\right) \propto E^n$ then the value of ' $n$ ' is

Accepted Answer

If $E$ is the kinetic energy of the proton, then
$\mathrm{E}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}$
where p is the momentum and m is the mass of proton
$	herefore \quad \mathrm{p}=\sqrt{2 \mathrm{mE}}$
$	herefore \quad \lambda_1=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$
Similarly, for a photon, $E=\frac{h c}{\lambda_2}$
$$
\begin{aligned}
	herefore\quad& \lambda_2=\frac{h c}{E} \
	herefore\quad& \frac{\lambda_1}{\lambda_2}=\frac{h}{\sqrt{2 \mathrm{mE}}} 	imes \frac{\mathrm{E}}{\mathrm{hc}}=\frac{1}{\mathrm{c}} \sqrt{\frac{\mathrm{E}}{2 \mathrm{~m}}} \
	herefore\quad& \frac{\lambda_1}{\lambda_2} \propto \mathrm{E}^{1 / 2} \Rightarrow \mathrm{n}=0.5
\end{aligned}$$

Answer

1

Answer

2

Answer

5

Answer

0.5

Dual Nature of Matter and Radiation

Kinetic energy of a proton is equal to energy E of a photon. Let ' \lambda_1 ' be the de-Broglie wavelength of proton and ' \lambda_2 ' be the wavelength of photon. If \left(\frac{\lambda_1}{\lambda_2}\right) \propto E^n then the value of ' n ' is

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