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

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

Given, initial velocity , `vecv=v_0hati`
The de Broglie wavelength associated with
`lamda_0=h/(mv_0)`
Electric field, `vecE=E_0hatj`
Force on electron due to electric field
`vecF=-evecE=-eE_0hatj`
Acceleration attained by the electron is
`veca=vecF/m=(-eE_0)/mhatj`
The acceleration acting on the electron is along negative y-axis.
Initial velocity along X-axis, `vecv_("xi")=v_0hati`
Initial velocity along Y-axis `vecv_("yi") =0`
Final velocity along X-axis after along t, `vecv_("xf") =vecv_("xi")=vecv_(0)hati`
(Since there is no acceleration along X-axis)
Final velocity along Y-axis after time t is
`vecu_(yf)=vecV_(yi)+vecat`
`=0+((-eE_0)/mhatj)t=(-eE_0)/mthatj`
Net velocity (magnitude) of electron after time t is
`|vecv_f|=sqrt(v_(yf)^2+v_(xf)^2)`
`=sqrt(((eE_0)/mt)^2+v_0^2)=v_0sqrt(1+(e^2E_0^2t^2)/(m^2v_0^2))`
The de Broglie wavelength associated with the electron at time t is
`lamda_t=h/(mv_f)=h/(mv_0sqrt(1+e^2E_0^2t^2//(m^2v_0^2)))`
`=lamda_0/(sqrt(1+e^2E_0^2t^2//m^2v_0^2))`