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, \[\frac{{{r}_{p}}}{{{p}_{d}}}=\sqrt{\frac{{{m}_{p}}}{{{m}_{d}}}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}\]
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