The kinetic energy of an electron gets triple | vedclass

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(C) The de-Broglie wavelength $\lambda$ of an electron is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's co

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$1/\sqrt{3}$

Vedclass · Answer

(C) The de-Broglie wavelength $\lambda$ of an electron is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass of the electron,and $K$ is the kinetic energy.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{K}}$.Let the initial kinetic energy be $K$ and the final kinetic energy be $K' = 3K$.The initial wavelength is $\lambda = \frac{h}{\sqrt{2mK}}$ and the final wavelength is $\lambda' = \frac{h}{\sqrt{2m(3K)}} = \frac{h}{\sqrt{3} \sqrt{2mK}}$.Therefore,the ratio of the wavelengths is $\frac{\lambda'}{\lambda} = \frac{1}{\sqrt{3}}$.Thus,the de-Broglie wavelength changes by a factor of $1/\sqrt{3}$.

The kinetic energy of an electron gets tripled,then the de-Broglie wavelength associated with it changes by a factor of:

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