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 debrogle wavelength `= (h)/(m_(0)v_(0))`
After any time `t = lamda = (h)/(m_(0)v_(0)+q_(0)E_(0)t)`
When `lamda` becomes half of the initial value :
`(h)/(2m_(0)v_(0))=(h)/(m_(0)v_(0)+q_(0)E_(0)t)`
`rArr m_(0)v_(0)=q_(0)E_(0)t rArr t = sqrt(3) (m_(0)v_(0))/(q_(0)E_(0))`.