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Electrons of mass m with de-Broglie wave...

Electrons of mass m with de-Broglie wavelength `lambda` fall on the target in an X-ray tube. The cut-off wavelength `(lambda_(0))` of the emitted X-ray is

A

` lambda_(0) = (2mc lambda^(2))/(h) `

B

`lambda_(0) = (2h)/(mc)`

C

`lambda_(0) = (2m^(2) c^(2) lambda^(3))/(h^(2))`

D

`lambda_(0) = lambda `

Text Solution

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The correct Answer is:
A
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