An electron (mass m) with an initial velocity `v=v_(0)hati` is in an electric field `E=E_(0)hatJ.If lamda_(0)=h//mv_(0),` it's de-Boroglie wavelength at time t is give by

Options

A. `lamda_0`
B. `lamda_0sqrt(1+(e^2E_0^2t^2)/(m^2v_0^2))`
C. `lamda_0/(sqrt(1+(e^2E_0^2t^2)/(m^2v_0^2)))`
D. `lamda_0/((1+(e^2E_0^2t^2)/(m^2v_0^2)))`

Answer

C

Explanation

Initial de-Broglie wavelength of electron
`lamda_0 = h/(mv_0)`
Force on electron in electric field, `vecF =-evecE=-eE_0hatj`
Acceleration of electron, `veca=vecF/m=(-eE_0hatj)/m `
It is acting along negative y-axis.
The initial velocity of electron along x-axis, `vecv_(x_0) = v_0hati`
Initial velocity of electron along y-axis, `vecv_(y_0)=0`
Velocity of electron after time t along x-axis, `v_x =v_0hati`
( `:.` there is no electron along x – axis)
Velocity of electron after time t along y-axis,
`vecv_y=0+(-(eE_0)/mhatj)t=-(eE_0)/m thatj`
Magnitude of velocity of electron after time t is
`v=sqrt(v_x^2+v_y^2)=sqrt(v_0^2+((-eE_0)/mt)^(2))implies v_0sqrt(1+(e^2E_0^et^2)/(m^2v_0^2))`
de-Broglie wavelength, `lama.=h/(mv)`
`implies h/(mv_0sqrt(1+e^2 E_0^2 t^2 //(m^2v_0^2)))`
`=(lamda_0)/(sqrt(1+e^2E_0^2t^2//m^2v_0^2))`