1.

An electron of mass m and charge e initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t, ignoring relativistic effect is :

Answer»

`-(H)/(eEt^(2))`
`-(eht)/(m)`
`(-mh)/(eEt^(2))`
`-(h)/(eE)`

SOLUTION :The ACCELERATION produced DUE to electric field is
`e=(eE)/(m)`
Now, `v=u+at+(eEt)/(m)`
de-Brogilie wavelength of an electrons is given by
`lambda=(h)/(mv)=(h)/(eEt)`
`:. (d lambda)/(DT)=-(h)/(-eEt^(2))`


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