Electrons are accelerated through a potential difference of 16 kV . If the potential difference is increased…
Electrons are accelerated through a potential difference of 16 kV . If the potential difference is increased to 64 kV , then de-Broglie wavelength associated with electron will
remain same.
becomes half.
becomes four time.
becomes quarter.
Solution
For an electron accelerated through potential V,
$\begin{array}{ll}
& \lambda \propto \frac{1}{\sqrt{V}} \text { i.e. } \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{V_2}{v_1}} \\
\therefore & \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{64}{16}}=2 \\
\therefore & \lambda_2=\frac{\lambda_1}{2}
\end{array}$