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Q.

A proton of energy 8 e V is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be

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a

4 eV

b

2 eV

c

8 eV

d

6 eV

answer is C.

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Detailed Solution

r=2mKqB  ⇒q ∝  mK ⇒  K  ∝  q2m⇒KαKp  = qαqp2 × mpmα ⇒ Kα8  = 2qpqp2 × mp4mp=1⇒  Kα =  8 eV
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