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If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is
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Verified Answer
The correct answer is:
more than that of a proton
Given,
$$
\begin{aligned}
\lambda &=\frac{h}{m v}=\frac{h}{\sqrt{2 m K E}} \\
K E &=\frac{h^{2}}{2 m \lambda^{2}}
\end{aligned}
$$
$$
\Rightarrow \quad \mathrm{KE}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}
$$
As $\lambda$ is same for both electron and proton
So, $\quad \mathrm{KE} \propto \frac{1}{\mathrm{~m}}$
Hence, kinetic energy will be maximum for particle with lesser mass, electron.
$$
\begin{aligned}
\lambda &=\frac{h}{m v}=\frac{h}{\sqrt{2 m K E}} \\
K E &=\frac{h^{2}}{2 m \lambda^{2}}
\end{aligned}
$$
$$
\Rightarrow \quad \mathrm{KE}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}
$$
As $\lambda$ is same for both electron and proton
So, $\quad \mathrm{KE} \propto \frac{1}{\mathrm{~m}}$
Hence, kinetic energy will be maximum for particle with lesser mass, electron.
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