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Step-by-Step Solution
Step 1: Identify the quantities and formulas
We have a singly ionized magnesium ion (mass number A = 24 ), which is accelerated to a kinetic energy of 5 keV (1 keV = 1000 \, \text{eV} ). The ion moves perpendicularly into a magnetic field of magnitude B = 0.5 \, \text{T} . We want to find the radius R of the circular path traced by the ion.
For a charged particle of mass m , charge q , and velocity v entering a uniform magnetic field B at right angles, the radius of the path is given by the formula:
R = \frac{mv}{qB}.
Since the ion is accelerated, we can relate its kinetic energy KE to velocity v using
KE = \frac{1}{2}mv^2,
which leads to
v = \sqrt{\frac{2 \times KE}{m}}.
Hence, we can write the radius in terms of m , KE , B , and q as:
R = \frac{\sqrt{2m \cdot KE}}{qB}.
Step 2: Convert and substitute the values
1. Mass of the magnesium ion ( m ): since A = 24 for magnesium, and the mass of 1 nucleon is approximately 1.67 \times 10^{-27} \, \text{kg} , the mass of the magnesium ion can be estimated as:
m \approx 24 \times 1.67 \times 10^{-27} \, \text{kg}.
2. Charge of a singly ionized magnesium ion ( q ): since it has lost one electron, q = e = 1.6 \times 10^{-19} \, \text{C}.
3. Kinetic energy ( KE ): 5 keV = 5 \times 1000 \, \text{eV} = 5000 \, \text{eV}. Since 1 eV = 1.6 \times 10^{-19} \, \text{J} ,
KE = 5000 \times 1.6 \times 10^{-19} = 8.0 \times 10^{-16} \, \text{J}.
4. Magnetic field ( B = 0.5 \, \text{T} ).
Step 3: Calculate the radius
Using
R = \frac{\sqrt{2m \cdot KE}}{qB},
substitute all the values:
m = 24 \times 1.67 \times 10^{-27} \, \text{kg},
\, KE = 8.0 \times 10^{-16} \, \text{J},
\, q = 1.6 \times 10^{-19} \, \text{C},
\, B = 0.5 \, \text{T}.
R = \frac{\sqrt{2 \times (24 \times 1.67 \times 10^{-27}) \times (8.0 \times 10^{-16})}}{(1.6 \times 10^{-19}) \times 0.5}.
Carrying out the calculation gives approximately:
R \approx 10 \, \text{cm}.
Step 4: State the final answer
The radius of the path of the singly ionized magnesium ion in the magnetic field is about 10 cm.
