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

An electron (mass m) with an initial velocity `vecv=v_(0)hati` is in an electric field `vecE=E_(0)hatj`. If `lambda_(0)=h//mv_(0)`. It's de-broglie wavelength at time t is given by

A

`lamda_(0)`

B

`lamda_(0) sqrt(1+ (e^(2) E_(0)^(2)t^(2))/(m^(2)v_(0)^(2)))`

C

`(lamda_(0))/(sqrt((1+ e^(2)E_(0)^(2) t^(2))/(m^(2)v_(0)^(2))))`

D

`(lamda_(0))/((1 + (e^(2) E_(0)^(2) t^(2))/(m^(2)v_(0)^(2))))`

Text Solution

Verified by Experts

The correct Answer is:
C
According to the problem de-Broglie wavelength of electron at time t= 0 is
`lamda_(0)= (h)/(mv_(0))`
Electrostatic force on electron in electric field is `vec(F)_(e )= -e vec(E )= -e E_(0) hat(j)`
The acceleration of electron, `vec(a)= (vec(F ))/(m)= -(e E_(0))/(m) hat(j)` It is acting along negative y-axis. The inital velocity of electron along x-axis. `v_(x_(0))= v_(0) hat(i)`
This component of velocity will remain constant as there is no force on electron in this direction. Now considering y-direction, initial velocity of electron along y-axis, `v_(y^(0))= 0`. Velocity of electron after time t along y-axis
`v_(y) = + (-(eE_(0))/(m)J) t= -(eE_(0))/(m) t hat(j)`
Magnitude of velocity of electron after time t is
`v= sqrt(v_(x)^(2) + v_(y)^(2))= sqrt(v_(0)^(2) + ((-E_(0))/(m)t)^(2))`
`rArr v_(0) sqrt(1 + (e^(2)E_(0)^(2) t^(2))/(m^(2)v_(0)^(2)))`
de-Broglie wavelength, `lamda_(0)= (h)/(mv_(0))`
`=(h)/(mv_(0) sqrt(1+ (e^(2)E_(0)^(2)t^(2))/(m^(2) v_(0)^(2))))= (lamda_(0))/(sqrt(1+ (e^(2) E_(0)^(2) t^(2))/(m^(2)v_(0)^(2))))`
`rArr lamda= (lamda_(0))/((1+ (e^(2)E_(0)^(2) t^(2))/(m^(2) v_(0)^(2)))`
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