In a mass spectrometer, singely ionized carbon-12 (normal carbon of atomic mass number 12) has a radius of curvature of 9 cm. What would be the radius of curvature of singely ionized carbon-14, assuming equal velocities?
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To find the radius of curvature of singly ionized carbon-14 in a mass spectrometer, we need to understand the relationship between the radius of curvature and the particle's mass and charge.
In a mass spectrometer, charged particles are accelerated in a magnetic field. The path of a charged particle in a magnetic field is bent due to the Lorentz force. The radius of curvature of the particle's path can be calculated using the following equation:
r = (m * v) / (z * B),
where:
- r is the radius of curvature,
- m is the mass of the particle,
- v is the velocity of the particle,
- z is the charge of the particle, and
- B is the strength of the magnetic field.
In this case, we are comparing singly ionized carbon-12 and carbon-14, which means that the charges of both ions are the same (z = 1). Additionally, the question states that the velocities of both carbon ions are equal.
Since the charge and velocity are the same for both carbon-12 and carbon-14, the only difference between them is their mass. Carbon-14 has a greater mass than carbon-12 because it contains two extra neutrons.
Therefore, we can conclude that the radius of curvature for carbon-14 would be greater than the radius of curvature for carbon-12. This is because a heavier mass requires a larger radius of curvature to maintain the same velocity.
However, we cannot calculate the exact value of the radius of curvature for carbon-14 without knowing the strength of the magnetic field or the velocity of the particles.