Question

Electrons accelerated by voltage $V$ are collimated and used in Young's double slit experiment. Now they are accelerated by voltage $4V$. If initially, fringe width was $\beta$, then new fringe width will be

• A. $2\beta$
• B. $3\beta$
• C. $4\beta$
• D. $\frac{\beta }{2}$

Answer 1

The correct option is D $\frac{\beta }{2}$

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$
$=\frac{h}{\sqrt{2mq\Delta V}}$

$\beta =\frac{D\lambda }{2}$
$\beta \propto \lambda \propto \frac{1}{\sqrt{\Delta V}}$
$\Delta V^{\prime }=4V$
$\therefore {\beta }^{{}^{\prime }}=\frac{\beta }{2}$