An electron of mass `m` with an initial velocity
`vec(v) = v_(0) hat`(i) `(v_(0) gt 0)` enters an electric field
`vec(E ) =- E_(0) hat (i)` `(E_(0) = constant gt 0)` at `t = 0` . If `lambda_(0)` is its de - Broglie wavelength initially, then its de - Broglie wavelength at time `t` is

Options

A. `lamda_0/((1+(eE_0)/(mv_0)t))`
B. `lamda_0 (1+(eE_0)/(mv_0)t)`
C. `lamda_0t`
D. `lamda_0`

Answer

A

Explanation

Here, `vecE=-vecE_0 hati` initial velocity `vecv=v_0 hati`
Force acting on electron due to electric field
`vecF=(-e) (-E_0hati)=eE_0hati`
Acceleration produced in the electron , `veca=vecF/m=(eE_0)/mhati`
Now, velocity of electron after time t,
Acceleration produced in the electron , `veca=(vecF)/m=(eE_0)/mhati`
Now, velocity of electron after time t,
`vecv_t=vecv+veca=(v_0+(eE_0t)/m)hati " or " |vecv_t|=v_0+(eE_0t)/m`
Now `lamda_t =h/(mv_t)=h/(m(v_0+(eE_0t)/m))=h/(m(1+(eE_0t)/m))`
`=h/((1+(eE_0t)/m))" "( :. lamda_0 =h/(mv_0))`