Solution:
de Broglie wavelength is \(lambda = \frac{h}{\sqrt{2mK}}\). If \(K' = 16K\), then \(lambda' = \frac{\lambda}{\sqrt{16}} = \frac{\lambda}{4}\). The percentage decrease is \(\frac{\lambda - \lambda/4}{\lambda} \times 100% = 75%\).
If the kinetic energy of a particle is increased to 16 times, the percentage decrease in de Broglie wavelength of particle is
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