A) \[1:4\]
B) \[1:\sqrt{2}\]
C) \[1:1\]
D) \[1:2\]
Correct Answer: B
Solution :
For a moving charge in a perpendicular magnetic field, \[\frac{m{{v}^{2}}}{r}=Bqv\] \[\Rightarrow \] \[r=\frac{mv}{Bq}=\frac{P}{Bq}\] or \[\frac{{{r}_{p}}}{{{r}_{d}}}=\frac{{{P}_{p}}}{{{P}_{d}}}\] ... (i) (as q is same for both) Also, momentum \[p=\sqrt{2\,mE}\] or \[\frac{{{P}_{p}}}{{{P}_{d}}}=\sqrt{\frac{{{m}_{p}}}{{{m}_{d}}}}\] ... (ii) From Eqs. (i) and (ii), we have, \[\frac{{{r}_{p}}}{{{r}_{d}}}=\sqrt{\frac{{{m}_{p}}}{{{m}_{d}}}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}\]
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