A particle of charge `q_(0)` and of mass `m_(0)` is projected along the `y`-axis at `t=0` from origin with a velocity `V_(0)`. If a uniform electric field `E_(0)` also exist along the `x`-axis, then the time at which debroglie wavelength of the particle becomes half of the initial value is:

Options

A. `(m_(0)v_(0))/(q_(0)E_(0))`
B. `2(m_(0)v_(0))/(q_(0)E_(0))`
C. `sqrt(3) (m_(0)v_(0))/(q_(0)E_(0))`
D. `3(m_(0)v_(0))/(q_(0)E_(0))`

Answer

C

Explanation

Initial de broglie wavelength `=h/(m_(0)v_(0))`
after any time t, `lambda=h/(sqrt((m_(0)v_(0))^(2)+(q_(0)E_(0)t)^(2)))`
When `lambda` becomes half of the initial value
`h/(2m_(0)v_(0))=h/(sqrt(m_(0)v_(0))^(2)+(q_(0)E_(0)t)^(2))=sqrt(3) m_(0)v_(0)=q_(0)E_(0)t rArr t=sqrt(3) (m_(0)v_(0))/(q_(0)E_(0))`