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. `lamda_(0) = (2mclamda^2)/(h)`
B. `lamda_(0) = (2h)/(mc)`
C. `lamda_(0) = (2m^2c^2lamda^3)/(h^2)`
D. `lamda_0 = lamda `

Answer

A

Explanation

Let K be kinetic energy of the incident electron.
Its linear momentum `p = sqrt(2mK)`
The de Broglie wavelength is related to the linear momentum as
`lamda = h/p = h/ (sqrt(2mK) )" or " K = h^2/(2mlamda^2)`
The cut - off wavelength of the emitted X - ray is related to the kinetic energy of incident electron as
`(hc)/(lamda_0) = K = (h^2)/(2lamda^2)implies lamda_(0) = (2mclamda^2)/h`