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Kinetic energy of the particle is E and ...

Kinetic energy of the particle is E and it's De-Broglie wavelength is `lambda`. On increasing it's KE by `delta E`, it's new De-Broglie wavelength becomes `lambda/2`. Then `delta E` is

A

4E

B

3E

C

2E

D

E

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The correct Answer is:
B
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