d
Kinetic energy gained by a charged particle accelerated by a potential \(V\) is \(qV\) \(KE = qV\)

\(\frac{ p ^{2}}{2 m }= qV \Rightarrow p =\sqrt{2 mqV }\)

\(p =\frac{ h }{\lambda}, \text { thus } \lambda=\frac{ h }{\sqrt{2 mqV }}\)

now \(\frac{\lambda_{ p }}{\lambda_{ d }}=\sqrt{\frac{ m _{ d } V _{ d }}{ m _{ p } V _{ p }}}\)

\(\Rightarrow \frac{1}{\sqrt{2}}=\sqrt{\frac{2 V_{d}}{V_{p}}} \Rightarrow \frac{V_{p}}{V_{d}}=4\)