Electron of mass m and intial velocity `vecV=v_(0)hati`
`(V_(0)gt0)` enter electric field `vecE=-E_(0)hati(E_(0))` constant at t=0 initial de-Broglie wavelength of electron is `lambda_(0)` then at time t=t its de-Broglie wavelength will be…..

Options

A. `lambda_(0)`
B. `(lambda_(0))/((1+(eE_(0))/(mV_(0))t))`
C. `lambda_(0)t`
D. `(1+(eE_(0))/(mV_(0))t)`

Answer

B

Explanation

`vecv=v_(0)hati,vecE=-E_(0)hati`
`therefore veca=(vecF)/(m)=(qvecE)/(m)`
`((-e)(-E_(0))hati)/(m)=(eE_(0))/(m)hati`….(1)
From equation of motion with constant acceleration,
v=u+at [From equation (1)]
`therefore v=v_(0)+(eE_(0))/(m)t`
Now,de-Broglie wavelength,
`lambda=(h)/(mv)=(h)/(m[v_(0)+(eE_(0))/(m)t])`
`(h)/(mv_(0)[1+(eE_(0))/(mv_(0))t])`
`=(lambda_(0))/(1+(eE_(0))/(mv_(0))t) [because (h)/(mv_(0))=lambda_(0)]`