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Q.

A monoenergetic electron beam with a de Broglie wavelength of x $Å$ is acceleratred till its wavelength is halved. By what factor is its kinetic energy changed?

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a

8

b

6

c

4

d

3

answer is C.

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Detailed Solution

$λ = \frac{h}{\sqrt{2} m E}$

$λ E =$ constant

So $λ_{1} \sqrt{E_{1}} = λ_{2} \sqrt{E_{2}}$

$λ_{2} / λ_{1} = \sqrt{E_{2} / E_{1}} = 2$

Therefore, $E_{2} / E_{1} = 4$

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A monoenergetic electron beam with a de Broglie wavelength of x Å is acceleratred till its wavelength is halved. By what factor is its kinetic energy changed?