A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume h=6.63 $ imes 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 imes 10^{-31} \mathrm{~kg}$ and $\mathrm{m}_{\mathrm{p}}=1836$ times $\mathrm{m}_{\mathrm{e}}$ )

Question

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume h=6.63 $	imes 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 	imes 10^{-31} \mathrm{~kg}$ and $\mathrm{m}_{\mathrm{p}}=1836$ times $\mathrm{m}_{\mathrm{e}}$ )

Aicharya · Accepted Answer

$\lambda$ is same for both $P=\frac{h}{\lambda}$ same for both $\mathrm{P}=\sqrt{2 \mathrm{mK}}$ Hence, $\begin{aligned} & \mathrm{K} \propto \frac{1}{\mathrm{~m}} \ & \Rightarrow \frac{\mathrm{KE}_{\mathrm{p}}}{\mathrm{KE}_{\mathrm{e}}}=\frac{\mathrm{m}_{\mathrm{e}}}{\mathrm{m}_{\mathrm{p}}}=\frac{1}{1836}\end{aligned}$

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies…

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