Atomic StructureHard
Question
An electron and a proton are accelerated through a potential V. If Pe and Pp are their momentum, then PP: Pe ratio is approximately equal to
Options
A.1: 1836
B.1: 1
C.1836: 1
D.43: 1
Solution
$eV = \frac{1}{2}mv^{2} = \frac{p^{2}}{2m} \Rightarrow p = \sqrt{2meV}$
$\therefore\frac{p_{p}}{p_{e}} = \sqrt{\frac{1836}{1}} = \frac{42.85}{1}$
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