The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ Natural carbon consists of two different isotopes (excluding $^{14} \mathrm{C},$ which is present in only trace amounts). The isotopes have different masses, which is due to different numbers of neutrons in the nucleus; however, the number of protons is the same, and subsequently the chemical properties are the same. The most abundant isotope has an atomic mass of \(12.00 \mathrm{u} .\) When natural carbon is placed in a mass spectrometer, two lines are formed on the photographic plate. The lines show that the more abundant isotope moved in a circle of radius \(15.0 \mathrm{cm},\) while the rarer isotope moved in a circle of radius \(15.6 \mathrm{cm} .\) What is the atomic mass of the rarer isotope? (The ions have the same charge and are accelerated through the same potential difference before entering the magnetic field.)

Step by step solution

01

Write the expression for centripetal force and magnetic force on the ions.

Centripetal force can be written as \(F_c = \frac{mv^2}{r}\), where m is the mass of the ion, v is its speed, and r is the radius of the circle formed in the spectrometer. The force exerted by the magnetic field on the ions can be written as \(F_b = qvB\), where q is the charge of the ion and B is the magnetic field strength. The centripetal force is provided by the magnetic force, so we have \(F_c = F_b\).

02

Set up the ratio between the masses and radii of the isotopes.

Since the forces experienced by both isotopes are the same, we can set up the following ratio: $$\frac{m_1}{m_2}=\frac{r_1}{r_2}= \frac{15.0 \mathrm{cm}}{15.6 \mathrm{cm}}$$

03

Solve for the mass of the rarer isotope.

We know the mass of the most abundant isotope, \(m_1 = 12.00 \, \mathrm{u}\). We can use the above ratio to find the mass of the rarer isotope, \(m_2\): $$m_2 = m_1 \cdot \frac{r_2}{r_1} = 12.00 \, \mathrm{u} \cdot \frac{15.6 \mathrm{cm}}{15.0 \mathrm{cm}}$$ Calculating \(m_2\): $$m_2 = 12.00 \cdot \frac{15.6}{15.0} = 12.43 \, \mathrm{u}$$ So, the atomic mass of the rarer isotope is \(12.43 \, \mathrm{u}\).

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