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. `sqrt3(m_(0)v_(0))/(q_(0)E_(0))`
D. `3(m_(0)v_(0))/(q_(0)E_(0))`

Answer

C

Explanation

`a=(E_(0)q_(0))/m,v=u+at,lambda=h/(mv)`.For de-Broglie's wavelength to become half of its initial value its speed is to be increased to `2v_(0)`
`v_(x)=0+(E_(0)q_(0))/mt,v_(y)=v_(0)` and `v^(2)=v_(x)^(2)+v_(y)^(2)`
i.e.`4v_(0)^(2)=v_(x)^(2)+v_(0)^(2) rArr v_(x)=sqrt3 v_(0) rArr t=sqrt3(m_(0)v_(0))/(q_(0)E_(0))`