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

Options

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`

Answer

A

Explanation

Let K be the kinetic energy of the incident electron. Its linear momentum `p=sqrt(2mK)`
The de-Broglie wavelength is related to the linear momentum as
`lambda = (h)/(p) = (h)/(sqrt(2mK)) ` or `K= (h^2)/(2 m lambda^2)`
The cut off wavelength of the emitted X-ray is related to the kinetic energy of incident electron as
`(hc)/(lambda_0) = K = (h^2)/(2m lambda^2) rArr lambda_0 = (2mc lambda^2)/(h)`