A) \[{{r}_{\alpha }}={{r}_{p}}={{r}_{d}}\]
B) \[{{r}_{\alpha }}={{r}_{p}}<{{r}_{d}}\]
C) \[{{r}_{\alpha }}>{{r}_{d}}>{{r}_{p}}\]
D) \[{{r}_{\alpha }}={{r}_{d}}>{{r}_{p}}\]
Correct Answer: B
Solution :
| [b] \[r=\frac{\sqrt{2mE}}{3q}\] |
| \[r\propto \frac{\sqrt{m}}{q}\] |
| \[{{r}_{p}}=k\frac{\sqrt{m}}{q}\] |
| \[{{r}_{D}}=k\frac{\sqrt{2m}}{q}\] |
| \[{{r}_{\alpha }}=k\frac{\sqrt{4m}}{2q}=\frac{k\sqrt{m}}{q}\] |
| \[\therefore \,\,\,{{r}_{p}}={{r}_{\alpha }}<{{r}_{d}}\]. |
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