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Electrons with de- Broglie wavelength la...

Electrons with de- Broglie wavelength `lambda` fall on the target in an X- rays tube . The cut off wavelength of the emitted X- rays is

A

0

B

`(hc)/(mc)`

C

`(2m^(2)c^(2)lambda^(2))/(h^(2))`

D

`(2mclambda^(2))/h`

Text Solution

Verified by Experts

The correct Answer is:
D
De - brogile wavelength
`lambda =h/p " " rArr P = h/lambda `
` :. ` Kinetic energy of electron `rArr E = (P^(2))/(2m) = (h^(2))/(2mlambda^(2))`
For cut-offf wavelength of emitted X - Ray
`E = (hc)/(lambda_(c ))`
`rArr (h^(2))/(2mlambda^(2))=(hc)/(lambda_(c))rArr lambda_(c ) =(2mclambda^(2))/h `
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