InterviewSolution
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InterviewSolution
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A proton and deuteron are accelerated by same potential difference.Find the ratio of their de-Broglie wavelengths. |
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Answer» The de Broglie wavelength `lambda` associated with same potential V is `lambda = (h)/(sqrt(2 meV))` `therefore " " lambda alpha (1)/(sqrtm)` [For same potential] Since alpha particle is `""_(2)^(4)He` and proton is `_(1)^(1)H` `therefore` mass of alpha particle = 4 times mass of proton `therefore " " alpha _("proton") alpha (1)/(sqrtM)` and `lambda_("alpha") alpha (1)/(sqrt4m)` `therefore (lambda_("proton"))/(lambda_("alpha")) = (1)/(sqrtm) xx (sqrt(4m))/(1) = (2)/(1) therefore lambda_("proton") : alpha_("alpha") = 2 : 1` |
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