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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, $\quad \lambda_{e}=\lambda_{p}$
Since, de-Broglie wavelength,
$$
\begin{aligned}
\lambda &=\frac{h}{m v}=\frac{h}{p}=\frac{h}{\sqrt{2 m K}} \\
\Rightarrow \quad \lambda &=\frac{h}{\sqrt{2 m K}}
\end{aligned}
$$
where, $K$ is kinetic energy.
$\begin{array}{ll}\therefore \quad \frac{\lambda_{e}}{\lambda_{p}} & =\sqrt{\frac{m_{p}}{m_{e}}} \cdot \sqrt{\frac{K_{p}}{K_{e}}} \\ \Rightarrow \quad \quad \frac{\lambda_{e}}{\lambda_{e}} & =\sqrt{\frac{m_{p}}{m_{e}}} \cdot \sqrt{\frac{K_{p}}{K_{e}}} \\ \Rightarrow \quad \text { [From Eq. (i)] }\end{array}$
$\Rightarrow \quad \sqrt{\frac{K_{e}}{K_{p}}}=\sqrt{\frac{m_{p}}{m_{e}}}$
Squaring both sides, we get
$$
\begin{array}{ll}
& \frac{K_{e}}{K_{p}}=\frac{m_{p}}{m_{e}} \\
\text { since } & m_{p}>m_{e} \\
\therefore & K_{e}>K_{p}
\end{array}
$$
Since, de-Broglie wavelength,
$$
\begin{aligned}
\lambda &=\frac{h}{m v}=\frac{h}{p}=\frac{h}{\sqrt{2 m K}} \\
\Rightarrow \quad \lambda &=\frac{h}{\sqrt{2 m K}}
\end{aligned}
$$
where, $K$ is kinetic energy.
$\begin{array}{ll}\therefore \quad \frac{\lambda_{e}}{\lambda_{p}} & =\sqrt{\frac{m_{p}}{m_{e}}} \cdot \sqrt{\frac{K_{p}}{K_{e}}} \\ \Rightarrow \quad \quad \frac{\lambda_{e}}{\lambda_{e}} & =\sqrt{\frac{m_{p}}{m_{e}}} \cdot \sqrt{\frac{K_{p}}{K_{e}}} \\ \Rightarrow \quad \text { [From Eq. (i)] }\end{array}$
$\Rightarrow \quad \sqrt{\frac{K_{e}}{K_{p}}}=\sqrt{\frac{m_{p}}{m_{e}}}$
Squaring both sides, we get
$$
\begin{array}{ll}
& \frac{K_{e}}{K_{p}}=\frac{m_{p}}{m_{e}} \\
\text { since } & m_{p}>m_{e} \\
\therefore & K_{e}>K_{p}
\end{array}
$$
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