An `alpha-`particle is describing a circle of radius 0.45 m in a field of magnetic induction of 1.2 T. Find its speed, frequency of rotation and kinetic energy. What potential difference will be required to accelerate the particle so as to give this much of the energy to it? The mass of `alpha-`particle is `6.8 xx 10^(-27) kg` and its charge is twice the charge of proton, i.e.,`3.2 xx 10^(-19) C`.

Answer

`2.6xx10^7 ms^-1 ; 9.2xx10^6s^-1 ; 14.3 MeV ; 7xx10^6V`.

Explanation

we know that
`F=evB=(mv^2)/r or v=(eBr)/m`
Subtituting the given values, we get
`v=((3.2xx10^-19)(1.2)(0.45))/((6.8xx10^-27))=2.6xx10^7ms^-1`
The frequency of rotation is given by
`f=v/(2pir)=(2.6xx10^7)/(2xx3.14xx0.45)=9.2xx10^6s^-1`
The kinetic energy of `alpha`-particle is given by
`K=1/2mv^2=1/26.8xx10^-27xx(2.6xx10^7)^2=2.3xx10^-12J`
`=(2.3xx10^-12)/(1.6xx10^-19)eV=14.3xx10^6eV=14.3MeV`