Question

Electrons with de - Brogli wavelengtyh `lambda` fall on the target in
an X-ray tube.The cut off wavelength of the emitted Xrays is
(a) `lambda_0 = (2mclambda^2)/(h)` (b)`lambda_0 = (2h)/(mc)`
(c ) `lambda_0 (2m^2 c^2 lambda^3)/(h^2)` (d)`lambda_0 = lambda`

Answer 1

Correct Answer - A::B::C::D
Momentum of bombarding electrons,`
`p = (h)/(lambda)`
`:. Kinetic energy of bombarding electrons,
`K = (p^2)/(2m) = (h^2)/(2mlambda^2)`
This is also maximum energy of X-ray photons.
Therefore, `(hc)/(lambda_0) = (h^2)/(2mlambda^2)
or `lambda_0 = (2mlambda^2 c)/(h)`
:. Correct option is (a).