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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

Vedclass · Accepted Answer

$75$

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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 $K_2 = 16K$.Then,the ratio of the wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}} = \sqrt{\frac{K}{16K}} = \sqrt{\frac{1}{16}} = \frac{1}{4} = 0.25$.So,$\lambda_2 = 0.25 \lambda_1$.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 the particle is increased by $16$ times,the percentage change in the de Broglie wavelength of the particle is ............$\%$

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