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The speed of a proton is a critical concept when dealing with problems that involve magnetic fields and forces. Knowing the speed helps us understand how a proton navigates through a magnetic field. In this scenario, the proton travels at an angle of \(23.0^{\circ}\) to the magnetic field. The formula used to determine the proton's speed originates from the magnetic force equation:\[ F = qvB \sin(\theta) \]Here:- \( F \) is the magnetic force (\(6.50 \times 10^{-17} \text{ N}\)).- \( q \) is the charge of the proton, approximately \(1.6 \times 10^{-19} \text{ C}\).- \( B \) is the magnetic field's strength (\(2.60 \text{ mT} = 2.60 \times 10^{-3} \text{ T}\)).- \( \theta \) is the angle (\(23.0^{\circ}\) in this case).To find the speed \( v \), we rearrange the equation:\[ v = \frac{F}{qB \sin(\theta)} \]Plugging in the known values and calculating the speed yields \( v \approx 6.53 \times 10^6 \text{ m/s} \). That means the proton is moving incredibly fast, showcasing how subatomic particles can travel at high speeds when influenced by a magnetic field.

Kinetic energy (KE) represents the energy that a particle possesses because of its motion. For the proton in question, its kinetic energy is directly related to its speed. The formula for kinetic energy is:\[ KE = \frac{1}{2} mv^2 \]Where:- \( m \) is the mass of the proton, approximately \(1.67 \times 10^{-27} \text{ kg}\).- \( v \) is the speed of the proton (\(6.53 \times 10^6 \text{ m/s}\)).Substituting these values into the kinetic energy equation gives:\[ KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times (6.53 \times 10^6)^2 \]After performing the calculation, we find that \( KE \approx 3.58 \times 10^{-13} \text{ J}\). This amount of kinetic energy may seem small, but given the proton's tiny mass, it reveals that even light particles can have significant energy due to their high speeds.

An electron-volt (eV) is a convenient unit for measuring energy on the scale of subatomic particles. One electron-volt is the amount of kinetic energy gained by a single electron when it is accelerated through an electric potential difference of one volt. This conversion is useful because it allows us to express particle energy in a more intuitive way than using joules.In our calculation, the proton's kinetic energy in joules \( (3.58 \times 10^{-13} \text{ J}) \) must be converted to electron-volts. The conversion factor is:1 eV = \( 1.6 \times 10^{-19} \text{ J} \).To convert, simply divide the kinetic energy in joules by this conversion factor:\[ KE \approx \frac{3.58 \times 10^{-13}}{1.6 \times 10^{-19}} \]Which results in approximately \( 2.24 \times 10^6 \text{ eV} \). Expressing kinetic energy in electron-volts provides a clearer understanding of the energy scales involved in subatomic physics.