A) \[1:\sqrt{2}:1\]
B) \[1:\sqrt{2}:\sqrt{2}\]
C) \[\sqrt{2}:1:1\]
D) \[\sqrt{2}:\sqrt{2}:1\]
Correct Answer: A
Solution :
At right angles to magnetic field, charged particles acquire circular path at which magnetic force provides necessary centripetal force to keep it moving on circular path. ie, \[qvB=\frac{m{{v}^{2}}}{r}\] \[\therefore \] \[r=\frac{mv}{qB}=\sqrt{\frac{2mE}{{{q}^{2}}{{B}^{2}}}}\] where\[E=KE\]of particle \[{{r}_{p}}=\sqrt{\frac{2mE}{{{e}^{2}}{{B}^{2}}}},\] \[{{r}_{d}}=\sqrt{\frac{2\times 2m\times E}{{{e}^{2}}{{B}^{2}}}}\] and \[{{r}_{\alpha }}=\sqrt{\frac{2\times 4m\times E}{{{(2e)}^{2}}{{B}^{2}}}}\] \[\therefore \] \[{{r}_{p}}:{{r}_{d}}:{{r}_{\alpha }}=1:\sqrt{2}:1\]
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