A) 4 MeV
B) 0.5 MeV
C) 1.5 MeV
D) 1 MeV
Correct Answer: D
Solution :
Radius in magnetic fields of circular orbit, \[R=\frac{mV}{qB}=\frac{\sqrt{2mB}}{qB}\] and total energy of a moving particle in a circular orbit, \[E=\frac{{{q}^{2}}{{B}^{2}}{{R}^{2}}}{2m}\] For a proton enter in a region of magnetic field \[{{E}_{1}}=\frac{{{e}^{2}}\times {{B}^{2}}\times {{R}^{2}}}{2\times {{m}_{p}}}\]is the mass of proton. Similarly for a \[\alpha \]-particle moves in a uniform magnetic field \[{{E}_{2}}\frac{{{(2e)}^{2}}\times {{B}^{2}}\times {{R}^{2}}}{2\times (4{{m}_{p}})}\] \[[\because \,{{m}_{\alpha }}\,=4{{m}_{p}}]\] Dividing Eq. (ii) by Eq. (i), we get \[\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{{{(2e)}^{2}}\times {{B}^{2}}\times {{R}^{2}}}{2\times (4{{m}_{p}})}\times \frac{2\times {{m}_{p}}}{{{e}^{2}}\times {{B}^{2}}\times {{R}^{2}}}\] \[\frac{{{E}_{2}}}{{{E}_{1}}}=1\Rightarrow {{E}_{2}}={{E}_{1}}=1\,\text{MeV}\]
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