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If the K.E. of free electron doubles, it...

If the K.E. of free electron doubles, its de-Broglie wavelength changes by the factor

A

`(1)/(2)`

B

2

C

`(1)/(sqrt(2))`

D

`sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
C

`lambda = (h)/(sqrt(2mK))`
When kinetic energy is doubled, `lambda. = (h)/(sqrt(2m xx 2K)) = (1)/(sqrt(2)) lambda`
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