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(B) The de-Broglie wavelength is given by the relation: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.From this relation,we can see that $\lambda

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half

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(B) The de-Broglie wavelength is given by the relation: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{KE}}$.Let the initial kinetic energy be $KE_1$ and the final kinetic energy be $KE_2 = 4 	imes KE_1$.Let the initial wavelength be $\lambda_1$ and the final wavelength be $\lambda_2$.Then,$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{KE_1}{KE_2}} = \sqrt{\frac{KE_1}{4 	imes KE_1}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,the new wavelength becomes half of the original wavelength.

If the kinetic energy of an electron is increased four times,the wavelength of the de-Broglie wave associated with it would become

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