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(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mK}}$, where $h$ is Planck's constant, $m$ is the mass of t

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$75$

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(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mK}}$, where $h$ is Planck's constant, $m$ is the mass of the particle, and $K$ is the kinetic energy.From this relation, we see that $\lambda \propto \frac{1}{\sqrt{K}}$.Let the initial kinetic energy be $K_1 = K$ and the final kinetic energy be $K_2 = 16K$.The ratio of the wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}} = \sqrt{\frac{K}{16K}} = \sqrt{\frac{1}{16}} = \frac{1}{4} = 0.25$.Thus, $\lambda_2 = 0.25\lambda_1$.The percentage change in wavelength is given by $\frac{\lambda_1 - \lambda_2}{\lambda_1} 	imes 100\%$.Substituting the values, we get $\frac{\lambda_1 - 0.25\lambda_1}{\lambda_1} 	imes 100\% = 0.75 	imes 100\% = 75\%$.

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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