b
The de broglie wavelength \(\lambda\) associated with the electrons is \(\lambda=\frac{1.227}{\sqrt{V}} n m\)

where \(V\) is the accelerating potential in volt.

\(\lambda \propto \frac{1}{\sqrt{V}}\)

\(\therefore \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{100 \times 10^3}{25 \times 10^3}}=2\)

\(\lambda_2=\frac{\lambda_1}{2}\)