A) \[{{R}_{e}}={{R}_{p}}\]
B) \[{{R}_{p}}={{R}_{d}}\]
C) \[{{R}_{d}}={{R}_{\alpha }}\]
D) \[{{R}_{p}}={{R}_{\alpha }}\]
Correct Answer: C
Solution :
By using \[r=\frac{mv}{qB}=\frac{v}{\left( \frac{q}{m} \right)B}\Rightarrow r\propto \ \frac{1}{\left( q/m \right)}\] \[\because \ {{\left( \frac{q}{m} \right)}_{{{e}^{-}}}}>{{\left( \frac{q}{m} \right)}_{{{p}^{+}}}}>\left\{ {{\left( \frac{q}{m} \right)}_{d}}={{\left( \frac{q}{m} \right)}_{\alpha }} \right\}\] \[\therefore \ {{R}_{d}}={{R}_{\alpha }}\]
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