A) 1 : 4
B) 4 : 1
C) 1 : 2
D) 1 : 1
Correct Answer: D
Solution :
Radius of circular path in a transverse magnetic field is given by \[r=\frac{m\upsilon }{Bq}\] \[=\frac{\sqrt{2mE}}{Bq}\] or \[\frac{mE}{{{q}^{2}}}=\text{cosntant}\] [\[\because \] B and r are constant for proton and\[\alpha \]particle] \[\therefore \] \[\frac{{{E}_{_{1}}}}{{{E}_{2}}}=\frac{{{q}_{1}}^{2}}{{{q}_{2}}^{2}}.\frac{{{m}_{2}}}{{{m}_{1}}}\] For proton, \[{{q}_{1}}=e,\,{{m}_{1}}=m\] For \[\alpha -\]particle; \[{{q}_{2}}=2E,\,{{m}_{2}}=4m\] Hence, \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{e}^{2}}}{4{{e}^{2}}}.\frac{4m}{m}\] \[=\frac{1}{1}=1:1\]
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