A proton, a deuteron and an - particle having the same K.E are moving in circular trajectory in a constant magnetic field. If $r_{p}, r_{d} a n d r_{α}$ denote respectively the radii of the trajectories of these particles then

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$r_{α} = r_{p} < r_{d}$

$r_{α} > r_{d} < r_{p}$

$r_{α} = r_{d} > r_{p}$

$r_{α} = r_{p} = r_{d}$

answer is A.

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Detailed Solution

$\begin{array}{l} \begin{array}{l} \underline{proton} & \underline{deuteron} & α - particle \\ m_{p} = 1.672 \times 10^{- 27} kg & m_{d} = 2 \times m_{p} & m_{d} = 4 m_{p} \\ q_{p} = 1.6 \times 10^{- 19} c & q_{d} = 1.6 \times 10^{- 19} c & q_{α} = 2 \times 1.6 \times 10^{- 19} \\ q_{p} & q_{d} = q_{p} & q_{α} = 2 q_{p} \\ {K.E}_{p} = K & {K.E}_{d} = K & {K.E}_{α} = K \\ r_{p} = \frac{m_{p} v}{{Bq}_{p}} & r_{d} = \frac{m_{d} v}{{Bq}_{d}} & r_{α} = \frac{m_{α} v}{{Bq}_{α}} \\ r_{p} = \frac{m_{p} v}{{Bq}_{p}} & r_{d} = \frac{2 \times m_{p} v}{{Bq}_{p}} & r_{α} = \frac{{4m}_{p} v}{{B2q}_{p}} \\ r_{d} = 2 (\frac{m_{p} v}{{Bq}_{p}}) & r_{α} = 2 (\frac{m_{p} v}{{Bq}_{p}}) \\ \begin{array}{l} If K . E is considered \end{array} & \begin{array}{l} r_{d} = 2 r_{p} \end{array} & \begin{array}{l} r_{α} = 2 r_{p} \end{array} \\ \underline{r_{p} = \frac{\sqrt{2 {Km}_{p}}}{Bqp}} & r_{d} = \frac{\sqrt{2 K ({2m}_{p})}}{Bqp} & r_{α} = \frac{\sqrt{2 K ({4m}_{p})}}{B (2qp)} \\ \begin{array}{l} r_{d} = \sqrt{2} \frac{\sqrt{2 K ({2m}_{p})}}{Bqp} \\ r_{d} = \sqrt{2} (r_{p}) \end{array} & \begin{array}{l} r_{α} = \frac{2 \sqrt{2 {Km}_{p}}}{2Bqp} \\ r_{α} = r_{p} \end{array} \end{array} \\ r_{p} = r_{α} < r_{d} \end{array}$