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An alpha-particle and a proton are accelerated from rest by the same potential, thenthe ratio of their de-Broglie wavelength is |
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Answer» Solution :`lambda = h/(sqrt(2qVm)) , lambda_(alpha) = h/(sqrt(2xx2xxV XX m_(alpha)))` `implieslambda_(alpha) = (h)/(sqrt(2xx2xx V xx 4))` `lambda _P =(h)/(sqrt(2xx1xxVxx m_p)) , lambda_p = (h)/(sqrt(2xx1xxV xx 1))` `(lambda_p)/(lambda_(alpha)) = sqrt((16)/(2)) implies(lambda_P)/(lambda_(alpha)) = 2sqrt2` |
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