Question

An x-ray tube operates at 40 kV. Suppose the electron converts 70% of its energy into a photon at each collision. Find the lowest three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.

• A. 76 pm, 205 pm, 364 pm
• B. 44 pm, 148 pm, 493 pm
• C. 36 pm, 112 pm, 256 pm
• D. 78 pm, 95 pm, 106 pm

Answer 1

The correct option is B 44 pm, 148 pm, 493 pm


The electron has an initial stock of 40 keV of energy since it has been accelerated through a potential difference of 40 kV . The energy lost in the first interaction $\Delta E_1$ is 70% of its present kinetic energy which is 40 keV , $\Delta E_1=70\%of40keV$
$=0.7\times 40keV=28keV$
The energy lost in the second interaction will be 70% of its current energy, i.e $12keV(40keV-28keV)$ ,
$\Delta E_2=70\%of12keV$
$=0.7\times 12keV=8.4keV$
Similarly, in the third interaction,
$\Delta E_3=70\%of(12-8.4)keV=70\%of3.6keV=2.52eV$
is lost.
Since these are the three highest losses of energy, it will correspond to the three lowest wavelengths $(E\propto \frac{1}{\lambda }).Applying\Delta E_{photon}=\frac{hc}{\lambda }andusinghc=1242eV.nm-$
${\lambda }_1=\frac{1242eV.nm}{\Delta E_1}=\frac{1242eV.nm}{28000eV}=44pm$,
${\lambda }_2=\frac{1242eV.nm}{\Delta E_2}=\frac{1242eV.nm}{8400eV}=148pm$,
${\lambda }_3=\frac{1242eV.nm}{\Delta E_3}=\frac{1242eV.nm}{2520eV}=493pm$.