Question

The de-Broglie wavelength of electron falling on the target in an X-ray tube is $\lambda$. The cut-off wavelength of the emitted X-ray is

• A. ${\lambda }_0=\frac{(mc\lambda {)}^2}{h}$
• B. ${\lambda }_0=\frac{m^{2}c\lambda }{h^2}$
• C. ${\lambda }_0=\frac{2mc{\lambda }^2}{h}$
• D. ${\lambda }_0=\frac{mc{\lambda }^2}{h^2}$

Answer 1

The correct option is C ${\lambda }_0=\frac{2mc{\lambda }^2}{h}$

The de-Broglie wavelength of a particle (let electron) of mass $m$, having kinetic energy $E$, is given by:

$\lambda =\frac{h}{\sqrt{2mE}}$

${\lambda }^2=\frac{h^2}{2mE}$

$E=\frac{h^2}{2m{\lambda }^2}$ ..................eq1

The cut-off wavelength of X-ray, emitted by the electron of kinetic energy $E$, is given by:

${\lambda }_0=\frac{hc}{E}$ ..................eq2

Putting the value of $E$, from eq1 to eq2, we get:

${\lambda }_0=\frac{hc}{h^2/2m{\lambda }^2}$

${\lambda }_0=\frac{2mc{\lambda }^2}{h}$