A) 1
B) 2
C) \[\frac{1}{\sqrt{2}}\]
D) \[\sqrt{2}\]
Correct Answer: D
Solution :
Given, \[{{K}_{\alpha }}={{K}_{d}}=K,\,{{q}_{\alpha }}=q={{q}_{d}},\,{{m}_{\alpha }}=2{{m}_{d}}\] For\[\alpha \]-particles \[{{r}_{1}}=\frac{\sqrt{2{{m}_{1}}K}}{Bq}\] For deuteron \[{{r}_{2}}=\frac{\sqrt{2{{m}_{2}}K}}{Bq}\] \[\therefore \] \[\frac{{{r}_{1}}}{_{2}}=\sqrt{2}\] \[\left( \because {{m}_{2}}=\frac{{{m}_{1}}}{2} \right)\]You need to login to perform this action.
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