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What will be the ratio of de - Broglie wavelengths of proton and `alpha` - particle of same energy ? |
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Answer» Correct Answer - A `lambda = (h)/(sqrt(2 m E)) rArr lambda prop (1)/(sqrt(m)) rArr (lambda_(p))/(lambda_(p)) sqrt((m_(a))/(m_(p))) = (2)/(1)` |
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