If the kinetic energy of a particle in motion | vedclass

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(B) Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2$. Given that the kinetic energy is decreased by $36 \%$,we have:$K_2

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2$. Given that the kinetic energy is decreased by $36 \%$,we have:$K_2 = K_1 - 0.36 K_1 = 0.64 K_1$.The de Broglie wavelength $\lambda$ is related to kinetic energy $K$ by the formula:$\lambda = \frac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \frac{1}{\sqrt{K}}$.Therefore,the ratio of the final wavelength $\lambda_2$ to the initial wavelength $\lambda_1$ is:$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}} = \sqrt{\frac{K_1}{0.64 K_1}} = \sqrt{\frac{1}{0.64}} = \frac{1}{0.8} = 1.25$.This implies $\lambda_2 = 1.25 \lambda_1$.The percentage increase in the de Broglie wavelength is:$\frac{\Delta \lambda}{\lambda_1} 	imes 100 = \frac{\lambda_2 - \lambda_1}{\lambda_1} 	imes 100 = (1.25 - 1) 	imes 100 = 25 \%$.

If the kinetic energy of a particle in motion is decreased by $36 \%$,the increase in de Broglie wavelength of the particle is (in $\%$)

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