1.

An electron (mass m)with an initial velocity vecv=v_(0)hati is in an electric field vecE=E_(0)hatj . If l,ambda_(0)=h..mv_(0).it's de-Broglie wavelength at time t is given by

Answer»

`(lambda_(0))/((1+(eE_(0))/(m)-(t)/(v_(0)))`
`lambda_(0)(1+(eE_(0)t)/(mv_(0)))`
`lambda_(0)`
`lambda_(0)t`

Solution :According to EQUATION,v=`v_(0)`+at at of UNIFORMLY acceleratd motion,
`v=v_(0)+((F)/(m))t`
`THEREFORE v=v_(0)+((eE_(0))/(m))t`
`therefore v=v_(0)(1+(eE_(0))/(mv_(0))t)`…..(1)
`therefore VECF=qvecE`
`therefore vecF=(-e)(-E_(0)hati)`
`therefore vecF=eE_(0)hati`
`therefore F=eE_(0)`
de-Broglie wavelength ,`lambda=(h)/(mv)implieslambdaprop(1)/(v)`
`therefore (lambda)/(lambda_(0))=(v_(0))/(v)` (where `lambda_(0)`=initial de-Broglie wavelength ,`lambda`=final de-broglie wavelength)
`therefore lambda=(lambda_(0)v_(0))/(v)`
`=(lambda_(0)v_(0))/(v_(0)(1+(eE_(0))/(mv_(0))t)` [From equation(1)]
`therefore lambda=(lambda_(0))/(1+(eE_(0))/(mv_(0))t)`


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