Lead 212/82Pb (atomic mass = 211.991871 u) is converted into bismuth 212/83Bi (atomic mass = 211.991255 u) via β- decay. What is the energy (in MeV) released in this process?
Q = (211.991871 – 211.991255) •1.66•10^-19•(3•10^8) = 9.2•10^-14 J =
=5.75•10^5 eV = 0.575 MeV
To calculate the energy released in this process, we can make use of the mass-energy equivalence principle given by Einstein's famous equation, E=mc^2.
The change in mass (Δm) during this radioactive decay is the difference between the initial mass (m_initial) and the final mass (m_final) of the nuclei involved. The energy released (ΔE) can be calculated using the formula:
ΔE = Δm * c^2
First, we need to calculate the change in mass (Δm), which is the difference between the initial and final masses.
Δm = m_initial - m_final
Given the atomic mass of lead 212/82Pb as 211.991871 u and the atomic mass of bismuth 212/83Bi as 211.991255 u, we can substitute these values into the above equation:
Δm = 211.991871 u - 211.991255 u
After subtracting the values, we get:
Δm = 0.000616 u
Now, we can calculate the energy released (ΔE) using the value of Δm:
ΔE = Δm * c^2
The speed of light (c) is approximately 2.998 x 10^8 meters per second. To convert atomic mass units (u) to kilograms (kg), we can use the conversion factor: 1 u = 1.66053906660 x 10^-27 kg.
Let's calculate the value of ΔE:
ΔE = (0.000616 u) * (1.66053906660 x 10^-27 kg/u) * (2.998 x 10^8 m/s)^2
After performing the calculation, we get the energy released (ΔE):
ΔE ≈ 9.3519 x 10^-14 Joules
To express this energy in MeV (mega-electron volts), we need to convert it using the energy conversion factor: 1 MeV = 1.6 x 10^-13 Joules.
Let's calculate the energy released in MeV:
ΔE_MeV = (ΔE / 1.6 x 10^-13) MeV
By substituting the value of ΔE, we get:
ΔE_MeV ≈ (9.3519 x 10^-14 J) / (1.6 x 10^-13 J/MeV)
After performing the calculation, we find that the energy released in this process is approximately:
ΔE_MeV ≈ 0.5844 MeV