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 :

Options

A. `-(h)/(eEt^(2))`
B. `-(eht)/(m)`
C. `(-mh)/(eEt^(2))`
D. `-(h)/(eE)`

Answer

A

Explanation

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))`