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. `lambda_(0)`
B. `lambda_(0)sqrt(1+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2)))`
C. `lambda_(0)/sqrt(1+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2)))`
D. `lambda_(0)/((1+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2))))`

Answer

3

Explanation

`V_(x)=V_(0), V_(y)=V_(0)+at=V_(0)+(eE_(0))/(m)t`
`rArrV=sqrt(V_(x)^(2)+V_(y)^(2))` Resultant velocity is `V=sqrt(V_(0)^(2)+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2)))lambda=(h)/(mv)`