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Chapter 38: Problem 46

Find the atomic mass of iridium-193, whose binding energy is 7.94 MeV/nucleon.

Short Answer

Expert verified
The atomic mass of iridium-193 is approximately \(192.96 \, \text{u}\).

Step by step solution

01

Calculating the Binding Energy

Find the total binding energy of the iridium-193 atom. The given binding energy is per nucleon, and a nucleon can be either a proton or a neutron. Iridium-193 has 193 nucleons. Multiply the given binding energy by the total number of nucleons: \( \text{Total binding energy} = 7.94 \, \text{MeV/nucleon} \times 193 \, \text{nucleons} \).
02

Convert the binding energy to J

Convert the total binding energy from MeV to Joules, using the conversion factor \(1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}\). This is important because you will be using \(E=mc^2\) in the next step, and \(E\) is in Joules.
03

Find the mass defect

Apply the formula \(E=mc^2\) to find the mass defect (\(m\)), where \(E\) is energy and \(c\) is the speed of light (\(2.998 \times 10^8 \, \text{m/s}\)): \( m =\frac{E}{c^2}\). The resulting value is the mass defect, in kg.
04

Convert mass defect to atomic mass units

Convert the mass defect from kg to atomic mass units (u) using the conversion factor \(1 \, \text{kg} = 6.02214076 脳 10^{26} \, \text{u}\). Subtract the mass defect from the total number of nucleons to find the atomic mass of iridium-193.
05

Round off the final answer

After performing all calculations, round off the final answer to only 2 decimal places.

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