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The kinetic energy of an electron gets tripled, then the de-Broglie wavelength associated with it changes by a factor
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The correct answer is:
$\frac{1}{\sqrt{3}}$
de-Broglie wavelength of an electron is given by
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$
or $\quad \lambda \propto \frac{1}{\sqrt{\mathrm{K}}}$
$\therefore \quad \frac{\lambda^{\prime}}{\lambda}=\frac{1}{\sqrt{3 \mathrm{~K}}} \frac{\sqrt{\mathrm{K}}}{1}=\frac{1}{\sqrt{3}}$
or $\quad \lambda^{\prime}=\frac{\lambda}{\sqrt{3}}$
Hence, de-Broglie wavelength will change by factor $\frac{1}{\sqrt{3}}$.
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$
or $\quad \lambda \propto \frac{1}{\sqrt{\mathrm{K}}}$
$\therefore \quad \frac{\lambda^{\prime}}{\lambda}=\frac{1}{\sqrt{3 \mathrm{~K}}} \frac{\sqrt{\mathrm{K}}}{1}=\frac{1}{\sqrt{3}}$
or $\quad \lambda^{\prime}=\frac{\lambda}{\sqrt{3}}$
Hence, de-Broglie wavelength will change by factor $\frac{1}{\sqrt{3}}$.
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