Question
The ratio of the $${\lambda _{\min }}$$ in a Coolidge tube to $${\lambda _{{\text{de Broglie}}}}$$ of the electrons striking the target depends on accelerating potential $$V$$ as
A.
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto \sqrt V $$
B.
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto V$$
C.
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto \frac{1}{{\sqrt V }}$$
D.
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto \frac{1}{V}.$$
Answer :
$$\frac{{{\lambda _{\min }}}}{{{\lambda _{{\text{de Broglie}}}}}} \propto \frac{1}{{\sqrt V }}$$
Solution :
$$\eqalign{
& {\lambda _{\min }} = \frac{{hc}}{{eV}} \cr
& {\text{and}}\,\,{\lambda _{{\text{de Broglie}}}} = \frac{h}{\rho } = \frac{h}{{\sqrt {2meV} }}. \cr} $$