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Electrons of mass m with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cutoff wavelength ($\lambda_0$) of the emitted X-ray is:
A
$\lambda_0 = \frac{2m^2c^2\lambda^3}{h^2}$
B
$\lambda_0 = \lambda$
C
$\lambda_0 = \frac{2mc\lambda^2}{h}$
D
$\lambda_0 = \frac{2h}{mc}$
Explanation
Entire electron KE converts to one X-ray photon at cut-off.
Detailed Solution
$\lambda = \frac{h}{p} \Rightarrow p = \frac{h}{\lambda}$; $E = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$
For the X-ray cut-off: $E = \frac{hc}{\lambda_0}$
$\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2} \Rightarrow \lambda_0 = \frac{2mc\lambda^2}{h}$
For the X-ray cut-off: $E = \frac{hc}{\lambda_0}$
$\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2} \Rightarrow \lambda_0 = \frac{2mc\lambda^2}{h}$