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

Answer

C

Explanation

According to the question.
`v= v_(0)hat(i)`
`E=-E_(0)hat(i)`

Thus, magnitude of force on the electron due to the electric field,
`|F|=q|E| rArrF=eE_(0)`
From Newton's second law of motion,
F=ma
`therefore F=ma = eE_(0) rArr F=eE_(0) rArr a=(eE_(0))/m`........(i)
or `a=((-e)(-E_(0)hati))/m=(eE_(0))/mhat(i)`
From first equation of motion,
v=u + at
Here, u(initial velocity) = `v_(0)`
`rArr v=v_(0)+(eE_(0))/mt`...............(ii) from Eq. (i)
Initial de-broglie wavelength of the electron is given as
`lambda_(0) = h/((mv_(0)))rArr h=lambda mv_(0)`............(iii)
After time t, de-Broglie wavelength is given as
`lambda = h/(mv)`
Substituting the value of v from Eq. (ii) , we get
`lambda=h/(m(v_(0)+(eE_(0))/mt)) = h/(mv_(0)[1+(eE_(0))/(mv_(0))t])`
`(lambdamv_(0))/(mv_(0)[1+(eE_(0))/(mv_(0))t])`
`=lambda_(0)/[1+(eE_(0))/(mv_(0))t]`
`therefore (lambdamv_(0))/(mv_(0)[1+(eE_(0))/(mv_(0))t])`
`=lambda_(0)/[1+(eE_(0))/(mv_(0))t]`