1.

An electron of mass m, when accelerated through a potential difference V, has de-Broglie wavelength lamda, the de-Broglie wavelength associated with a protom of mass M and accelerated through the same potential difference V will be

Answer»

<P>`lamda(m)/(M)`
`lamdasqrt((m)/(M))`
`lamdasqrt((M)/(m))`
`(lamdaM)/(m)`

Solution :As per relation: `lamda=(h)/(SQRT(2mqV))` for same POTENTIAL V, we have
`(lamda_(p))/(lamda_(c))=(lamda_(p))/(lamda)=sqrt((m)/(M)) or lamda_(p)=lamdasqrt((m)/(M))`.


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