Question
Electrons with energy $$80\,keV$$ are incident on the tungsten target of an X-ray tube. $$K$$-shell electrons of tungsten have $$72.5\,keV$$ energy. X-rays emitted by the tube contain only
A.
a continuous X-ray spectrum (Bremsstrahlung) with a
minimum wavelength of $$0.155\mathop {\text{A}}\limits^ \circ $$
B.
a continuous X-ray spectrum (Bremsstrahlung) with all wavelengths
C.
the characteristic X-ray spectrum of tungsten
D.
a continuous X-ray spectrum (Bremsstrahlung) with a
minimum wavelength of $$0.155\mathop {\text{A}}\limits^ \circ $$ and the characteristic X-ray spectrum of tungsten.
Answer :
a continuous X-ray spectrum (Bremsstrahlung) with a
minimum wavelength of $$0.155\mathop {\text{A}}\limits^ \circ $$ and the characteristic X-ray spectrum of tungsten.
Solution :
KEY CONCEPT :
$$\eqalign{
& {\lambda _{\min }} = \frac{{hc}}{E} \cr
& \therefore {\lambda _{\min }} = \frac{{12400}}{{80 \times {{10}^3}}}\mathop {\text{A}}\limits^ \circ = 0.155\mathop {\text{A}}\limits^ \circ \cr} $$
Energy of incident electrons is greater than the ionization energy of electrons in $$K$$-shell, the $$K$$-shell electrons will be knocked off. Hence, characteristic X-ray spectrum will be obtained.