A singly ionized magnesium atom $$(A = 24)$$ ion is accelerated to kinetic energy $$5$$ keV, and is projected perpendicularly into a magnetic field $$B$$ of the magnitude $$0.5$$ T. The radius of path formed will be ______ cm.

We need to find the radius of the circular path of a singly ionized magnesium ion in a magnetic field. The mass number, $$A = 24$$, so mass $$m = 24 \times 1.67 \times 10^{-27}$$ kg $$= 24 \times 1.67 \times 10^{-27}$$ kg; the kinetic energy, $$KE = 5$$ keV $$= 5 \times 10^3 \times 1.6 \times 10^{-19}$$ J $$= 8 \times 10^{-16}$$ J; the magnetic field, $$B = 0.5$$ T; and the charge, $$q = 1.6 \times 10^{-19}$$ C (singly ionized).

Since $$KE = \frac{1}{2}mv^2$$, it follows that $$v = \sqrt{\frac{2 \times KE}{m}}$$.

The radius of circular motion in a magnetic field is given by $$r = \frac{mv}{qB}$$. Substituting $$v = \sqrt{\frac{2 \times KE}{m}}$$ gives

$$r = \frac{m}{qB} \sqrt{\frac{2 \times KE}{m}} = \frac{\sqrt{2m \times KE}}{qB}$$

Substituting the values:

$$m = 24 \times 1.67 \times 10^{-27} = 4.008 \times 10^{-26} \text{ kg}$$

$$2m \times KE = 2 \times 4.008 \times 10^{-26} \times 8 \times 10^{-16}$$

$$= 64.128 \times 10^{-42} = 6.4128 \times 10^{-41}$$

$$\sqrt{2m \times KE} = \sqrt{6.4128 \times 10^{-41}} = 8.008 \times 10^{-21}$$

Now calculate $$r$$:

$$r = \frac{8.008 \times 10^{-21}}{1.6 \times 10^{-19} \times 0.5}$$

$$r = \frac{8.008 \times 10^{-21}}{8 \times 10^{-20}}$$

$$r = \frac{8.008}{80} = 0.1001 \text{ m}$$

$$r \approx 0.1 \text{ m} = 10 \text{ cm}$$

Hence, the radius of the path is 10 cm.