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A particle of +e charge is analyzed with a mass spectrometer that has a magnetic field B = 0.050 T . If the particle has a mass of m=2.0×10−27 kg and a speed of 3.0×105 m/s , what is the radius of the path?

User Fhtuft
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Final answer:

The radius of the path for a charged particle in a mass spectrometer subject to a magnetic field can be calculated with the formula R = mv/(qB), where m is the mass, v is the velocity, q is the charge, and B is the magnetic field strength. Hence, the radius of the path is 0.075 m.

Step-by-step explanation:

The student is asking about the radius of the circular path taken by a charged particle in a mass spectrometer when subjected to a magnetic field. The magnetic field exerts a force on the charged particle, causing it to move in a circular path. The radius of this path can be calculated using the relationship R = mv/(qB), where m is the mass of the particle, v is the speed of the particle, q is the charge of the particle, and B is the magnetic field strength.

Given that the mass m of the particle is 2.0×10⁻²⁷ kg, the speed v is 3.0×10⁵ m/s, the charge q is the fundamental charge +e (approximately 1.6×10⁻¹⁹ C), and the magnetic field B is 0.050 T, we can calculate the radius of the path.

Substituting the values into the formula:
R = (2.0×10⁻²⁷ kg × 3.0×10⁵ m/s) / ((1.6×10⁻¹⁹ C) × 0.050 T)
Computing the values, we get the radius R.

R = 0.075

Hence, the radius of the path is 0.075 m.

User Kowal
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