An electron is moving with an initial velocity `vec(v)=v_(0)hat(i)` and is in a magnetic field `vec(B)=B_(0)hat(j)`. The it's de Broglie wavelength

Options

A. `((2pimv_(0))/e((B_(x)^(2))/(B_(0)^(3))),(2pimv_(0))/e((B_(x)B_(y))/(B_(0)^(3))))`
B. `((2pimv_(0))/e((B_(x)B_(y))/(B_(0)^(3))),(2pimv_(0))/e((B_(x)B_(y))/(B_(0)^(3))))`
C. `((2pimv_(0))/e((B_(x)^(2))/(B_(0)^(3))),(2pimv_(0))/e((B_(y)^(2))/(B_(0)^(3))))`
D. `((2pimv_(0))/e((B_(x)B_(y))/(B_(0)^(3))),(2pimv_(0))/e((B_(y)^(2))/(B_(0)^(3))))`

Answer

D

Explanation

The force due to the electric field and the force due to gravity cancel each other at all time instants, so we need not analyse the effects of these two forces.
Let the angle made by the magnetic field vector with the Y-axis be `theta`. So `(B_(x))/(B_(y))=tan theta`
The magnetic field is given by `vecB=(B_(0)sin theta)hati+(B_(0)cos theta)hatj`
The electron will follow a helical path. The component of velocity of the electron parallel to the field will remain unchanged, and the components of its velocity in a plane perpendicular to the field will oscillate.
At `t=0` component of velocity parallel to field
`v_(y)=v_(0)cos theta`
And, component of velocity perpendicular to field,
`v_(_|__)=v_(0)sin theta`
So radius of the helical path `r=(mv_(_|_))/(eB_(0))`
And time perido `T=(2pim)/(eB_(0))`
The electron will return to the X-Y plane when it has completed one full rotation,
i.e. at `t=T=(2pim)/(eB_(0))`
During this time, the displacement of the electron (in a direction parallel to the field),
`L=v_(y)((2pim)/(eB_(0)))=(2pimv_(0)cos theta)/(eB_(0))`
, coordinates of the point where the electron intersects the X-Y plane are
` (L sin theta, L cos theta)`
i.e. `((2pi mv_(0)sin theta cos theta)/(eB_(0)), (2pi m v_(0)cos^(2) theta)/(eB_(0)))`, i.e. `(((2pi m v_(0))/e)((B_(x)B_(y))/(B_(0)^(3))),((2pimv_(0))/e)((B_(y)^(2))/(B_(0)^(3))))`