An electron gun fires an electron beam of kinetic energy E with a very small divergence angle `theta`. To focus all the electrons to one point a magnetic field induction 'B' is applied along the direction of propagation of the electrons. If the mass and charge on a electron are m and q respectively, then minimum distance x (shown in figure) at which the electrons focus is `etaxx(sqrt(mE))/(qB)` then value of `eta` is:

Answer

8.87

Explanation

Since angle between velocity vector and magnetic field direction `theta` is in between `0^(@)` and `90^(@)` so the path of electron will be helical and electrons will focus again on distances equal to integral multiplication of pitch distances.
Required minimum distance `x=`Pitch
`=v cos theta xxT=v cos theta xx (2pim)/(qB)=sqrt((2E)/m)xxcos thetaxx(2pim)/(qB)=2sqrt(2)pi cos theta xx(sqrt(mE))/(qB)[:' theta` is small `:.cos theta ~~1]`