A) \[\frac{1}{2}\]
B) \[\frac{1}{4}\]
C) 1
D) 4
Correct Answer: A
Solution :
: \[\alpha -\]particle \[{{=}_{2}}H{{e}^{4}}\] \[\therefore \]charge\[=2e\] Proton\[{{=}_{1}}{{H}^{1}}\] \[\therefore \] charge\[=e\] Momentum\[={{m}_{\alpha }}{{v}_{\alpha }}={{m}_{p}}{{v}_{p}}\] For circular motion in magnetic field, \[Bev=\frac{m{{v}^{2}}}{R}\] or \[BeR=mv\] \[\therefore \]\[B{{e}_{\alpha }}{{R}_{\alpha }}=B{{e}_{p}}{{R}_{p}}\]or \[\frac{{{R}_{\alpha }}}{{{R}_{p}}}=\frac{{{e}_{P}}}{{{e}_{\alpha }}}=\frac{e}{2e}=\frac{1}{2}\] \[\therefore \] \[\frac{{{R}_{\alpha }}}{{{R}_{p}}}=\frac{1}{2}\].
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