A particle has mass m. If its kinetic energy | vedclass

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(B) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the particle.We know that kinetic energy $K$

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Becomes half

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(B) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the particle.We know that kinetic energy $K$ is related to momentum $p$ by the formula $K = \frac{p^2}{2m}$,which implies $p = \sqrt{2mK}$.Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mK}}$.Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2 = 4K_1$.The initial wavelength is $\lambda_1 = \frac{h}{\sqrt{2mK_1}}$ and the final wavelength is $\lambda_2 = \frac{h}{\sqrt{2mK_2}}$.Taking the ratio: $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}} = \sqrt{\frac{K_1}{4K_1}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,the new de Broglie wavelength $\lambda_2$ becomes half of the initial wavelength $\lambda_1$.

$A$ particle has mass $m$. If its kinetic energy is increased to four times its initial value,what will be the change in its de Broglie wavelength?

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