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

C

Explanation

Initial de Broglie wavelength of electron,`lambda_(0)=h/(mv_(0))`
Force on electron in electric field,`vecF=-evecE=-eE_(0)hatj`
Acceleration of electron, `veca=vecF/m=-(eE_(0))/mhatj`.It is acting along negative `y`-axis.The initial velocity of electron along `x`-axis `vecv_(x_0)=v_(0)hati`.Initial velocity of electron along `y`-axis `vecv_(y_0)`.
Velocity of electron after time `t` along `x`-axis,`vecv_(x)=v_(0)hati`.
`(because` there in no acceleration of electron along `x`-axis )
Velocity of electron after time `t` along `y`-axis, `vecv_(y)=0+(-(eE_(0))/mhatj)t=-(eE_(0))/mthatj`
Magnitude of velocity of electron after time `t` is `|vecv|=sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(v_(0)^(2)+((-eE_(0))/my)^(2))=v_(0)sqrt(l+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2))`
de Broglie wavelength associated with electron at time `t` is `lambda=h/(mv)=h/(mv_(0)sqrt(1+(e^(2)E_(0)^(2)t^(2))/((m^(2)v_(0)^(2)))))=lambda_(0)/sqrt(1+(e^(2)E_(0)^(2)t^(2))/((m^(2)v_(0)^(2))))`