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A proton and an alpha particle, both initially at rest, are accelerated so as to have the same kinetic energy. What is the ratio of their de-Broglie wavelength ? |
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Answer» <P> SOLUTION :de-Broglie wavelength,`lambda = (h)/(p) = (h)/(sqrt(2 mK))` i.e. `lambda prop (1)/(sqrt(m)) ""[m_(ALPHA) = 4 m_(p)]` `(lambda_(p))/(lambda_(alpha)) = sqrt((m_(alpha))/(m_(p))) = sqrt((4 m_(p))/(m_(p))) = sqrt((4)/(1)) = (2)/(1)` `lambda_(p) : lambda_(alpha) = 2 : 1` |
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