Electrons of mass m with de–Broglie wavelength λ fall on the target in an X–ray tube. The cutoff wavelength (λ0) of the emitted X

Electrons of mass m with de–Broglie wavelength λ fall on the target in an X–ray tube. The cutoff wavelength (λ₀) of the emitted X–ray is

$\lambda_0 = \lambda$

$\lambda_0 = \dfrac{2mc\lambda^2}{h}$

$\lambda_0 = \dfrac{2h}{mc}$

$\lambda_0 = \dfrac{2m^2c^2\lambda^3}{h^2}$

Explanation:

K.E. of electrons = $\dfrac{P^2}{2m}= \dfrac{\left(\dfrac{h}{\lambda}\right)^2}{2m} = \dfrac{h^2}{2m\lambda^2}$

So maximum energy of photon will also be this much.

$\dfrac{hc}{\lambda_0} = \dfrac{h^2}{2m\lambda^2}$

$\lambda_0 = \dfrac{2mc\lambda^2}{h}$