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An electron (mass m) with an initial vel...

An electron (mass m) with an initial velocity `v=v_(0)hat(i)(v_(0)gt0)` is in an electric field `E=-E_(0)hat(l)(E_(0)="constant"gt0)`. Its de-Broglie wavelength at time t is given by

A

`(lamda_(0))/([1+ (e E_(0))/(m) (t)/(v_(0))])`

B

`lamda_(0)[1+ (e E_(0)t)/(mv_(0)]`

C

`lamda_(0)`

D

`lamda_(0)t`

Text Solution

Verified by Experts

The correct Answer is:
A
The wave associated with moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity According to de-Broglie theory, the wavelength of de-Broglie wave is given by
`lamda= (h)/(p)= (h)/(mv)= (h)/(sqrt(2mE))`. As inital velocity of the electron is `v_(0) hat(i)`, then initial de-Broglie wavelength of electron, `lamda= (h)/(mv_(0))` ...(i)
Electrostatic force on electron in electric field is, `vec(F)_(e )= -e vec(E )= -e [-E_(0) hat(i) ]= e E z_(0)hat(i)`
Acceleration of electron, `vec(a)= (vec(F ))/(m)= (e E_(0)hat(j))/(m)`
Velocity of the electron after time t, `vec(v)= v_(0) hat(i) + ((e E_(0) hat(i))/(m)) t= (v_(0) + (eE_(0))/(m)t) hat(i)`
`rArr vec(v)= v_(0) (1+ (eE_(0))/(mv_(0))) hat(i)`
de-Broglie wavelength associated with electron at time t is `lamda= (h)/(mv)`
`rArr lamda = (h)/(m [v_(0) (1+ (eE_(0))/(mv_(0))t)])= ((h)/(mv_(0)))/((1+ (eE_(0))/(mv_(0))t))`
`rArr lamda= (lamda_(0))/([1+ (eE_(0))/(mv_(0))t])` As `lamda= (h)/(mv_(0))`
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