An electron of mass \(m_{e}\) and a proton of mass \(m_{p}\) are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is

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Answer & Solution

Correct Answer

(A) \(\sqrt{\frac{m_{p}}{m_{E}}}\)

Step-by-step Solution

Detailed explanation

\(\text { Kinetic Energy }=e \Delta \,V\)

\(\lambda_{e}=\frac{h}{\sqrt{2 m_{e}(e \Delta V)}}\)

\(\lambda_{p}=\frac{h}{\sqrt{2 m_{p}(e \Delta V)}}\)

\(\Rightarrow \frac{\lambda_{e}}{\lambda_{p}}=\sqrt{\frac{m_{p}}{m_{e}}}\)

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