A neutral atom of atomic mass number 100 which is stationary at the origin in gravity free space emits an `alpha`-particle (A) in z-direction. The product ion is P. A uniform magnetic field exists in the x-direction. Disregard the electromagnetic interaction between A and P. If the angle of rotation of A after which A and P will meet for the first time is `(npi)/(25)` radians, what is the value of n?

Answer

`48`

Explanation

`A` and `P` will have the same momentum in magnitude and they will move in opposite directions.They will move in the circle of same radius and the same centre but in opposite directions.If they meet after time `t` then
`omega_(A)t+omega_(P)t=2pi`
`rArrt=(2pi)/(omega_(A)+omega_(P))=(2pi)/((2e.B)/(4m)+(2eB)/((A-4m)))`
`t=(4(A-4)mpi)/(eBA),0_(A)=omega_(A)t=(2eB)/(4m)xx(4m(A-4))/(eBA)`
`(2(A-4)pi)/A=48/25pi rArr n=48`