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An alpha-particle and a proton are accelerated through the same potential difference. Find the ratio of their de-Broglie wavelengths. |
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Answer» Solution :We know that if a prticle of mass m and charge q is accelerated through a potential V, its de-Broglie wavelength is given by `lamda=(lamda)/(mv)=(H)/(p)=(h)/(sqrt(2mK))=(h)/(sqrt(2qmV))` Where K is the kinetic enegy of charge particle. We know that `q_a=2q_(p) and m_(a)=4m_(p)` SINCE accelerating potential V is same, hence the RATIO of de-Broglie.s wavelengths is given by `(lamda_(p))/(lamda_(alpha))=sqrt((q_(alpha)*m_(alpha))/(q_(p)*m_(p)))=sqrt(((2q_(p))(4m_(p)))/(q_(p)*m_(p)))=2sqrt(2) implies lamda_(p)=2sqrt(2)lamda_(alpha)`. |
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