InterviewSolution
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InterviewSolution
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A particle of mass `M` at rest dacays into two particle of masses `m_(1)` and `m_(2) `having non zero velocity . The ratio of the de Broglie wavelength . The ratio of the de Broglie wavelength of the particle `lambda , _(1) // lambda_(2)` isA. `m_(1)//m_(2)`B. `m_(2)//m_(1)`C. `1.0`D. `sqrt(m_(2))// sqrt(m_(1))` |
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Answer» Correct Answer - C By law of conservation of momentum `0 = m_(1) vec(v_(1)) + m_(2) vec(v_(2)) rArr m_(1) vec (v_(1)) = - m_(2) vec(v_(2))` `- ve` sign indicates that both the particles are moving in opposite direction. Now de - Broglie wavelengths `lambda_(1) = (h)/(m_(1) v_(1))` and `lambda_(2) = (h)/(m_(2) v_(2)), :. (lambda_(1))/(lambda_(2)) = (m_(2) v_(2))/(m_(1) v_(1)) = 1` |
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