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Electrons with de Broglie wavelength lam...

Electrons with de Broglie wavelength `lamda` fall on the target in an X ray tube. The cut off wavelength `(lamda_(0))` of the emitted X rays is

A

`lamda_(0)=(2mclamda^(2))/(h)`

B

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

C

`lamda_(0)=(2m^(2)c^(2)lamda^(3))/(h^(2))`

D

`lamda_(0)=lamda`

Text Solution

Verified by Experts

The correct Answer is:
A
`{:(lamda_(0)=(hc)/(KE_(e))),(lamda_(0)=(hc)/(h^(2)//2mlamda^(2))):|{:(lamda=(h)/(sqrt(2mKE_(e)))),(KE_(e)=(h^(2))/(2mlamda^(2))):}`
`lamda_(0)=(2mc)/(h)lamda^(2)`
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