A positive, singly ionized atom of mass number $$A_M$$ is accelerated from rest by the voltage 192 V. Thereafter, it enters a rectangular region of width $$w$$ with magnetic field $$\vec{B}_0 = 0.1\hat{k}$$ Tesla, as shown in the figure. The ion finally hits a detector at the distance $$x$$ below its starting trajectory.

[Given: Mass of neutron/proton $$= (5/3) \times 10^{-27}$$ kg, charge of the electron $$= 1.6 \times 10^{-19}$$ C.]

Which of the following option(s) is(are) correct?

Using kinetic energy from potential difference: $$K = qV$$

Using path radius in a magnetic field: $$R = \frac{\sqrt{2mK}}{qB_0} = \frac{1}{B_0}\sqrt{\frac{2mV}{q}}$$

$$R = \frac{1}{0.1}\sqrt{\frac{2 \times A_M \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}} = 10\sqrt{A_M \times 400 \times 10^{-8}} = 0.02\sqrt{A_M}\text{ m} = 2\sqrt{A_M}\text{ cm}$$

Using geometry of the semicircular path back to the detector plane:

$$x = 2R = 4\sqrt{A_M}\text{ cm}$$

Evaluating $$x$$ for $$H^+$$ ion ($$A_M = 1$$): $$x = 4\sqrt{1} = 4\text{ cm}$$

Evaluating $$x$$ for $$A_M = 144$$: $$x = 4\sqrt{144} = 4 \times 12 = 48\text{ cm}$$

Evaluating detector limits for $$1 \le A_M \le 196$$:

$$x_{\text{min}} = 4\sqrt{1} = 4\text{ cm} \implies x_0 = 4\text{ cm}$$

$$x_{\text{max}} = 4\sqrt{196} = 4 \times 14 = 56\text{ cm} \implies x_1 = 56\text{ cm}$$

$$\text{Height } (x_1 - x_0) = 56 - 4 = 52\text{ cm}$$

Minimum field width $$w$$ to complete the semicircle for $$A_M = 196$$:

$$R_{\text{max}} = 2\sqrt{196} = 28\text{ cm} \implies w_{\text{min}} = R_{\text{max}} = 28\text{ cm}$$

Answer: Option (A), Option (B)