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Chapter 26: Problem 43

Radon decays as follows: $^{222} \mathrm{Rn} \rightarrow^{218} \mathrm{Po}+\alpha .$ The mass of the radon-222 nucleus is 221.97039 u, the mass of the polonium- 218 nucleus is \(217.96289 \mathrm{u},\) and the mass of the alpha particle is 4.00151 u. How much energy is released in the decay? \(\left(1 \mathrm{u}=931.494 \mathrm{MeV} / \mathrm{c}^{2} .\right)\)

Short Answer

Expert verified
Answer: The energy released in the decay of Radon-222 is approximately 5.582 MeV.

Step by step solution

01

Identify the initial and final masses

In the decay process, Radon-222 decays into Polonium-218 and an alpha particle. The initial mass is the mass of Radon-222, and the final masses are the masses of Polonium-218 and the alpha particle. We have: Initial mass (Radon-222): \(m_{Rn} = 221.97039\,\text{u}\) Final mass (Polonium-218): \(m_{Po} = 217.96289\,\text{u}\) Final mass (alpha particle): \(m_{\alpha} = 4.00151\,\text{u}\)
02

Find the mass difference

We can find the mass difference by subtracting the final masses from the initial mass: Mass difference, \(\Delta m = m_{Rn} - (m_{Po} + m_{\alpha})\) Substituting the given values, we get: \(\Delta m = 221.97039\,\text{u} - (217.96289\,\text{u} + 4.00151\,\text{u}) = 0.00599\,\text{u}\)
03

Convert mass difference to energy

We can convert the mass difference into energy using the mass-energy equivalence formula, which is given as: \(E = \Delta m \cdot c^2\) Here, \(E\) is the energy released, \(\Delta m\) is the mass difference, and \(c\) is the speed of light in a vacuum. Since we're given that 1 u is equal to 931.494 MeV/\(c^2\), we can substitute this value for the mass difference to obtain the energy in MeV: \(E = 0.00599\,\text{u} \cdot (931.494\,\text{MeV}/\text{c}^2) = 5.58169\,\text{MeV}\)
04

Round the result and conclude

Finally, let's round our result to a reasonable number of significant figures, since the given masses are given to six decimal places: Energy released: \(E \approx 5.582\,\text{MeV}\) So, the energy released in the decay of Radon-222 is approximately 5.582 MeV.

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