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?
(A) 12.3
(B) 10.5
(C) 7.71
(D) 3.58
mv^2/r=mv^2/r
m12/r12=m14/r14
solve for r14
r14=r12*m14/m12
Don't understand this...can u break it down
To calculate the radius of curvature of singly ionized carbon-14, we can use the formula for the centripetal force:
F = (mv²) / r
Where:
- F is the centripetal force acting on the ion
- m is the mass of the ion
- v is the velocity of the ion
- r is the radius of curvature
In this case, we are assuming that the velocities of both carbon-12 and carbon-14 ions are equal, so we can use this information to solve for the radius of curvature of carbon-14.
Let's first determine the mass of carbon-14. Carbon-12 and carbon-14 have atomic mass numbers of 12 and 14, respectively. Since we are dealing with singly ionized ions, their charges would be +1. Therefore, the mass of carbon-14 would be:
mass of carbon-14 = (atomic mass number) x (mass of proton)
mass of carbon-14 = 14 x (mass of proton)
Next, we will equate the centripetal force of carbon-14 with carbon-12 since they have the same velocity:
(mass of carbon-12) x (velocity of carbon-12)² / (radius of carbon-12) = (mass of carbon-14) x (velocity of carbon-14)² / (radius of carbon-14)
We know the mass of carbon-12 is 12 times the mass of the proton. The velocity of carbon-12 is equal to the velocity of carbon-14, so we can cancel these terms:
(12 x mass of proton)² / (radius of carbon-12) = (mass of carbon-14)² / (radius of carbon-14)
Simplifying, we get:
(radius of carbon-12 / radius of carbon-14) = (mass of carbon-14 / mass of carbon-12)²
Plugging in the given radius of carbon-12 (9 cm) and rearranging the equation, we get:
radius of carbon-14 = (mass of carbon-14 / mass of carbon-12)² x radius of carbon-12
Let's calculate this value using the given atomic mass numbers:
radius of carbon-14 = (14 / 12)² x 9 cm
radius of carbon-14 ≈ 10.5 cm
Therefore, the radius of curvature of singly ionized carbon-14, assuming equal velocities, is approximately 10.5 cm.
So, the correct answer is (B) 10.5.