The situation is show in
When a charged particle with charge `q` is accelerated through a potential difference V volt, then
`1/2mv^2=qV` …..(i)
or `v=sqrt(((2qV)/(m)))`…(i)
`alpha`-particle in magnetic field moves in a circle of radius `R` which is given by
`R=(mv)/(qB)` or `R=1/B sqrt(((2mV)/(q)))`....(ii)
The change in direction of `alpha`-particle `(theta)` from . is given by
`sin theta=1/R=lBsqrt((q/(2mV))`
Here, `l=0.1m ,B=0.1tesla, V=10^4`Volt
`q=2e=1.6xx10^-19=3.2xx10^-19C`
and `m=6.4xx10^-27kg`
`:. sin theta=0.1xx0.1xx sqrt(((3.2xx10^-19)/(2xx6.4xx10^-27xx10^4)))=1/2`

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