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  • A proton and an alpha particle both enter a region of uniform magnetie field B, moving at right angles to field B. If the radius of circular orbits for both the particles is equal and the kine

A proton and an alpha particle both enter a region of uniform magnetie field B, moving at right angles to field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV the energy acquired by the alpha particle will be:

Option: 1

0.5 MeV


Option: 2

1 MeV


Option: 3

1.5 MeV


Option: 4

4 MeV


Answers (1)

best_answer

\text { As we know, } F=q v B=\frac{m v^2}{R}

\\ \therefore R=\frac{m V}{q B}=\frac{\sqrt{2 m(K E)}}{q B}\\ since\ R\ is\ same\ so,\ K E \propto \frac{q^2}{m}

Therefore K.E of a particle

=\frac{q^2}{m}=\frac{(2)^2}{4}=1 \mathrm{MeV} .

Posted by

Pankaj Sanodiya

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