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An alpha- particle and a photon are accelerated from rest through the same potential difference V. Find the ratio of de Broglie wavelength associated with them.

Answer»

Solution :`K.E. = ( 1)/( 2 ) m upsilon^(2) = QV` or
`upsilon= sqrt((2qV)/(m))`
de Broglie WAVELENGTH,` LAMBDA = ( h )/( mv)`
For same POTENTIAL difference,
`( lambda_(ALPHA))/( lambda_(p))= sqrt((m_(p)q_(p))/( m_(alpha)q_(alpha)))=sqrt((m_(p)e)/( 4m_(p) 2e)) = (1)/( 2sqrt( 2))`


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