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An electron of mass m with an initial ve...

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

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
(c ) : Initial de Broglie wavelength of electron, `lamda_(0)=(h)/(mv_(0))`
Force on electron in electric field, `vec(F)=-evec(E)=-eE_(0)hat(j)`
Acceleration of electron, `vec(a)=(vec(F))/(m)=-(eE_(0))/(m)hat(j)`
It is acting along negative along x-axis `vec(v_(x0))=v_(0)hat(i)`.
Initial velocity of electron along y-axis `vec(v_(y0))=v_(0)hat(i)`
`( :.` there is no acceleration of electron along x-axis)
Velocity of electron after time t y- axis,
`vec(v_(y))=0+(-(eE_(0))/(m)hat(j))t=-(eE_(0))/(m)that(j)`
Magnitude of velocity of elecron after time t is
`|vec(v)|=sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(v_(0)^(2)+((-eE_(0))/(m)t)^(2))=v_(0)sqrt(1+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2)))`
de Broglie wavelength associated with electron at time t is
`lamda=(h)/(mv)=(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))))`
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