Question
Electrons of mass $$m$$ with de-Broglie wavelength $$\lambda $$ fall on the target in an X-ray tube. The cut-off wavelength $$\left( {{\lambda _0}} \right)$$ of the emitted X-ray is
A.
$${\lambda _0} = \frac{{2mc{\lambda ^2}}}{h}$$
B.
$${\lambda _0} = \frac{{2h}}{{mc}}$$
C.
$${\lambda _0} = \frac{{2{m^2}{c^2}{\lambda ^3}}}{{{h^2}}}$$
D.
$${\lambda _0} = \lambda $$
Answer :
$${\lambda _0} = \frac{{2mc{\lambda ^2}}}{h}$$
Solution :
Cut-off wavelength occurs when incoming electron looses its complete energy in collision. This energy appears in the form of X-rays.
Given, mass of electrons = $$m$$
de-Broglie wavelength = $$\lambda $$
So, kinetic energy of electron $$ = \frac{{{p^2}}}{{2m}}$$
$$ = \frac{{{{\left( {\frac{h}{\lambda }} \right)}^2}}}{{2m}} = \frac{{{h^2}}}{{2m{\lambda ^2}}}$$
Now, maximum energy of photon can be given by
$$\eqalign{
& E = \frac{{hc}}{{{\lambda _0}}} = \frac{{{h^2}}}{{2m{\lambda ^2}}} \cr
& \Rightarrow {\lambda _0} = \frac{{hc \times 2{\lambda ^2}.m}}{{{h^2}}} \cr
& = \frac{{2mc{\lambda ^2}}}{h} \cr} $$