A) 2MeV
B) 1 MeV
C) 0.5MeV
D) 4 MeV
Correct Answer: B
Solution :
[b] Radius of the circular path of a charged particle in uniform magnet field \[R=\frac{mv}{qB}=\frac{m}{Bq}\sqrt{\frac{2k}{m}}=\sqrt{\frac{2mk}{Bq}}\] For proton mass = m charge = q For a-particle mass = 4m charge = 2q \[{{R}_{\alpha }}=\frac{\sqrt{2\,(4m)k}}{B\,(2q)}=\frac{\sqrt{2m{{k}_{\alpha }}}}{Bq}\] \[\frac{{{R}_{p}}}{{{R}_{\alpha }}}=\sqrt{\frac{{{k}_{p}}}{{{k}_{\alpha }}}}\] As RP=R (given) \[{{K}_{\alpha }}={{K}_{p}}=1MeV\]You need to login to perform this action.
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