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Question: A proton and an electron are accelerated by the same potential difference. Let l<sub>e</sub> and l<s...

A proton and an electron are accelerated by the same potential difference. Let le and lp denote the de-Broglie wavelengths of the electron and the proton respectively.

A

le = lp

B

le<lp

C

le>lp

D

The relation between le and lp depends on the accelerating potential difference

Answer

le>lp

Explanation

Solution

le = h2meqV\frac{h}{\sqrt{2m_{e}qV}}, lp = h2mpqV\frac{h}{\sqrt{2m_{p}qV}}

̃ λeλp\frac{\lambda_{e}}{\lambda_{p}}= mpme\sqrt{\frac{m_{p}}{m_{e}}} > 1 ̃ le > lp