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Chapter 29: Problem 16

What is the mass of an \(^{16} \mathrm{O}\) atom in units of $\mathrm{MeV} / c^{2} ?\( (1 \)\mathrm{MeV} / \mathrm{c}^{2}$ is the mass of a particle with rest energy \(1 \mathrm{MeV} .)\)

Short Answer

Expert verified
Answer: The mass of the \(^{16} \mathrm{O}\) atom in units of \(\mathrm{MeV}/c^2\) is approximately \(16.663 \; \frac{MeV}{c^2}\).

Step by step solution

01

Convert the atomic mass into kilograms

Since the \(^{16} \mathrm{O}\) atom has a mass number of 16, its mass is approximately 16 atomic mass units (amu). 1 amu can be converted to kilograms using the conversion factor 1 amu = \(1.66054 \times 10^{-27}\) kg. The mass of the \(^{16} \mathrm{O}\) atom in kg can be calculated as: \(m_{O} =\) 16 amu \(\times\) \(1.66054 \times 10^{-27} \; \frac{kg}{amu}\) = \(2.656864 \times 10^{-26} \; kg\)
02

Convert mass into its energy equivalent using \(E=mc^2\)

Now, we will use Einstein's famous equation, \(E=mc^2\), to convert the mass of the Oxygen atom into energy. The speed of light, \(c\), is \(3 \times 10^8 \ m/s\). The energy equivalent of the \(^{16} \mathrm{O}\) atom in Joules is: \(E = m_{O} c^2 = (2.656864 \times 10^{-26} \; kg) \times (3 \times 10^8 \;\frac{m}{s})^2 = 2.390581 \times 10^{-10} \; J\)
03

Convert the energy into Mega Electron Volts (MeV)

Now, we need to convert the energy from Joules to Mega Electron Volts. The conversion factor is 1 MeV = \(1.60219 \times 10^{-13}\) J. The energy equivalent of the \(^{16} \mathrm{O}\) atom in MeV is: \(E_{MeV} = \frac{2.390581 \times 10^{-10} \; J}{1.60219 \times 10^{-13} \;\frac{J}{MeV}} = 1493.409 \; MeV\)
04

Calculate mass in units of \(\mathrm{MeV}/\mathrm{c^2}\)

Finally, we can convert the energy equivalent of the \(^{16} \mathrm{O}\) atom into mass units of \(\mathrm{MeV}/\mathrm{c^2}\). \(\frac{E_{MeV}}{\mathrm{c^2}} = \frac{1493.409 \;MeV}{(3 \times 10^8 \;\frac{m}{s})^2} = 16.663 \; \frac{MeV}{c^2}\) So, the mass of an \(^{16} \mathrm{O}\) atom in units of \(\mathrm{MeV}/c^2\) is approximately \(16.663 \; \frac{MeV}{c^2}\).

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