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The ratio between the de Broglie wavelengthassociated with protons, accelerated through a potential of 512 V and that of alpha particles accelerated through a potential of X volts is found to be one. Find the value of X. |
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Answer» Solution :de-Broglie WAVELENGTH of accelerated CHARGE particle `lambda = (h)/(sqrt(2mqV))` `lambda prop (h)/(sqrt(mq V))` Ratio of wavelength of proton and `ALPHA`-particle. `(lambda_(p))/(lambda_(alpha)) = sqrt((m_(alpha) q_(alpha) V_(alpha))/(m_(p)q_(p) V_(p))) = sqrt(((m_(alpha))/(m_(p)))((q_(alpha))/(q_(p)))((V_(alpha))/(V_(p))))` Here, `(m_(alpha))/(m_(p)) = 4, (q_(alpha))/(q_(p)) = 2, (V_(alpha))/(V_(p)) = (X)/(512), (lambda_(p))/(lambda_(alpha)) = 1` `1 = sqrt(4 xx 2 xx ((X)/(512))) = sqrt((X)/(64)) =(X)/(64)` X = 64 V |
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