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If the kinetic energy of a particle is increased by 16 times, the percentage change in the de Broglie wavelength of the particle is
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$75 \%$
$\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m E}} \quad$ or $\quad \lambda \propto \frac{1}{\sqrt{E}}$
$\therefore \frac{\lambda^{\prime}}{\lambda}=\sqrt{\frac{E}{E^{\prime}}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$
$\%$ change in wavelength $=\left(\frac{\lambda-\lambda^{\prime}}{\lambda}\right) \times 100$
$=\left(1-\frac{\lambda^{\prime}}{\lambda}\right) \times 100=\left(1-\frac{1}{4}\right) \times 100=75 \%$
$\therefore \frac{\lambda^{\prime}}{\lambda}=\sqrt{\frac{E}{E^{\prime}}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$
$\%$ change in wavelength $=\left(\frac{\lambda-\lambda^{\prime}}{\lambda}\right) \times 100$
$=\left(1-\frac{\lambda^{\prime}}{\lambda}\right) \times 100=\left(1-\frac{1}{4}\right) \times 100=75 \%$
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