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A deuteron and an alpha particl are accelerated with the same potential. Which one of the two hasgreater value of de Broglie wavelength associated with it and i |
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Answer» Solution :(i) Using de-Broglie wavelength formula, the dueteron and alpha particle are accelerated with same potential. So, boh their velocities are same. `lambda = (h)/(sqrt(2 mV_(0) Q)), lambda prop (1)/(sqrt(mq))` So, `lambda_(due) prop (1)/(sqrt(2m_(d) q_(d))) and lambda_(alpha) prop (1)/(sqrt(8m_(alpha)q_(alpha)))` `(lambda_(alpha))/(lambda_(d)) = (sqrt(2m_(d)q_(q)))/(sqrt(8m_(alpha) q_(alpha))) = sqrt((2)/(8)) = sqrt((1)/(4)) = (1)/(2)` `lambda_(d) = 2 lambda_(alpha)` (ii) For same potential of ACCELERATION, KE is directly proportional to the .q. Charge of duetron is +e Charge of alpha is +2e So, `K_(d) = (K_(alpha))/(2)` Charge of alpha particle is more than the duetron. |
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