Electrons with the de Broglie wavelength λ fall on the target in an X ray tube. The cut off wavelength $(λ_0)$ of the emitted X rays is:
Answer & explanation
Correct answer: option 1
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
$λ=\frac{h}{p}=\frac{h}{\sqrt{2mK}}$ or $K=\frac{h^2}{2mλ^2}$
The cut off wavelength of the emitted X rays is related to the kinetic energy of the incident electron as
$\frac{hc}{λ_0}=K=\frac{h^2}{2mλ^2}$ or $λ_0=\frac{2mcλ^2}{h}$