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.

<p>To calculate the radius of the path formed by a singly ionized magnesium atom in a magnetic field, the following formula is used:</p> <p>$ R = \frac{mv}{qB} $</p> <p>This can be rewritten using kinetic energy (KE):</p> <p>$ R = \frac{\sqrt{2m \cdot KE}}{qB} $</p> <p>Here's how we calculate it for a magnesium ion:</p> <p><p><strong>Atomic Mass (A)</strong>: 24</p></p> <p><p><strong>Mass of a Nucleon</strong>: $1.67 \times 10^{-27} $ kg (approximate mass of a proton or neutron)</p></p> <p><p><strong>Charge (q)</strong>: $1.6 \times 10^{-19} $ C (since the magnesium ion is singly ionized)</p></p> <p><p><strong>Magnetic Field (B)</strong>: 0.5 T</p></p> <p><p><strong>Kinetic Energy (KE)</strong>: 5 keV = $5 \times 1.6 \times 10^{-16} $ J</p></p> <p>Plug in these values to find the radius:</p> <p>$ R = \frac{\sqrt{2 \times 24 \times 1.67 \times 10^{-27} \times 5 \times 1.6 \times 10^{-16}}}{1.6 \times 10^{-19} \times 0.5} $</p> <p>This simplifies to:</p> <p>$ R = 10.009 \, \text{cm} \approx 10 \, \text{cm} $</p> <p>Thus, the radius of the path is approximately 10 cm.</p>