Q147 / 180 · KCET 2020 · Physics
A cyclotron is used to accelerate protons $\left({ }_{1}^{1} \mathrm{H}\right)$ deuterons $\left({ }_{1}^{2} \mathrm{H}\right)$ and $\alpha$-particles $\left({ }_{2}^{4} \mathrm{He}\right)$. While exiting under similar conditions, the minimum KE is gained by
- A$\alpha$-particles
- Bprotons
- Correct answerCdeuterons
- DSame for all
Solution
When a charged particle is accelerated by a potential difference $V$ volts, then the radius of the trajectory of the path of the particle is given as,
$\begin{aligned}
r &=\frac{\sqrt{2 K m}}{B q}=\frac{\sqrt{2 q V m}}{B q} \\
\Rightarrow \quad K &=\frac{1}{2} \frac{B^{2} q^{2} r^{2}}{m} \text { or } K \propto \frac{q^{2}}{m} ...(i)
\end{aligned}$
For protons $\left({ }_{1}^{1} \mathrm{H}\right)$, deuterons $\left({ }_{1}^{2} \mathrm{H}\right)$ and $\alpha$-particles $\left({ }_{2}^{4} \mathrm{He}\right)$,
$\begin{aligned} q_{p}: q_{d}: q_{\alpha} &=1: 1: 2 ...(ii)\\ m_{p}: m_{d}: m_{\alpha} &=1: 2: 4 ...(iii)\end{aligned}$
From Eqs. (i), (ii) and (iii), we can write
$\Rightarrow \quad K_{p}: K_{d}: K_{\alpha}=1: \frac{1}{4}: 1$
Thus, minimum kinetic energy is gained by deuterons.
Hence, the correct answer is option: 'C'.