A singly ionized magnesium atom (A=24) ion is | vedclass

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(D) Given: Mass of magnesium ion $m = 24 	imes 1.66 	imes 10^{-27} \, kg \approx 3.984 	imes 10^{-26} \, kg$. Kinetic energy $K = 5 \, keV = 5000 \

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$10$

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(D) Given: Mass of magnesium ion $m = 24 	imes 1.66 	imes 10^{-27} \, kg \approx 3.984 	imes 10^{-26} \, kg$. Kinetic energy $K = 5 \, keV = 5000 	imes 1.6 	imes 10^{-19} \, J = 8 	imes 10^{-16} \, J$. Charge $q = 1.6 	imes 10^{-19} \, C$. Magnetic field $B = 0.5 \, T$. The radius of the path is given by $R = \frac{\sqrt{2mK}}{qB}$. Substituting the values: $R = \frac{\sqrt{2 	imes 3.984 	imes 10^{-26} 	imes 8 	imes 10^{-16}}}{1.6 	imes 10^{-19} 	imes 0.5}$. $R = \frac{\sqrt{63.744 	imes 10^{-42}}}{0.8 	imes 10^{-19}} = \frac{7.984 	imes 10^{-21}}{0.8 	imes 10^{-19}} \approx 9.98 	imes 10^{-2} \, m$. Converting to centimeters: $R \approx 9.98 	imes 10^{-2} 	imes 100 \, cm \approx 10 \, cm$.

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

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