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The kinetic energy of an electron get tripled then the de-Broglie wavelength associated with it changes by a factor
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{3}}$
de-Broglie wavelength of an electron
$$
\begin{array}{l}
\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}} \text { or } \lambda \propto \frac{1}{\sqrt{\mathrm{K}}} \\
\therefore \frac{\lambda^{\prime}}{\lambda}=\frac{1}{\sqrt{3 \mathrm{~K}}}=\frac{\sqrt{\mathrm{K}}}{1}=\frac{1}{\sqrt{3}}
\end{array}
$$
or $\lambda^{\prime}=\frac{\lambda}{\sqrt{3}}$
i.e. de-Broglie wavelength will change by
factor $\frac{1}{\sqrt{3}}$.
$$
\begin{array}{l}
\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}} \text { or } \lambda \propto \frac{1}{\sqrt{\mathrm{K}}} \\
\therefore \frac{\lambda^{\prime}}{\lambda}=\frac{1}{\sqrt{3 \mathrm{~K}}}=\frac{\sqrt{\mathrm{K}}}{1}=\frac{1}{\sqrt{3}}
\end{array}
$$
or $\lambda^{\prime}=\frac{\lambda}{\sqrt{3}}$
i.e. de-Broglie wavelength will change by
factor $\frac{1}{\sqrt{3}}$.
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