A) \[{{r}_{\alpha }}={{r}_{p}}<{{r}_{d}}\]
B) \[{{r}_{\alpha }}>{{r}_{d}}>{{r}_{p}}\]
C) \[{{r}_{\alpha }}={{r}_{d}}>{{r}_{p}}\]
D) \[{{r}_{p}}={{r}_{d}}={{r}_{\alpha }}\]
Correct Answer: A
Solution :
Given that \[{{K}_{p}}={{K}_{d}}={{K}_{\alpha }}\]= K (say) We know that mp = m, md = 2m and \[{{m}_{\alpha }}=4m\]and qp = e, qd = e and \[{{q}_{\alpha }}=2e\] Further\[r=\frac{\sqrt{2mK}}{qB}\]Þ\[{{r}_{p}}=\frac{\sqrt{2mK}}{eB}\], \[{{r}_{d}}=\frac{\sqrt{2\left( 2m \right)K}}{eB}=\sqrt{2}{{r}_{p}}\] and \[{{r}_{\alpha }}=\frac{\sqrt{2\left( 4m \right)K}}{\left( 2e \right)B}={{r}_{p}}\]. Hence \[{{r}_{\alpha }}={{r}_{p}}<{{r}_{d}}\]
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