Question

A proton accelerated through a potential V has de-Broglie wavelength lamda. Then, the de-Broglie wavelengthof an alpha-particle, when accelerated through the same potential V is

Answer 1

Answer:

de broglie wavelength inversely proportional to

$\sqrt{mq}$

so wave length of alpha particle

$lambda \times 2 \sqrt{2}$

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