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Chapter 29: Problem 69

Radon gas (Rn) is produced by the \(\alpha\) decay of radium 226 88 28 A How many neutrons and how many protons are present in the nucleus of the isotope of Rn produced by this decay? (b) In the air in an average size room in a student basement apartment in Ithaca, NY, there are about \(10^{7}\) Rn nuclei. The Rn nucleus itself is radioactive; it too decays by emitting an \(\alpha\) particle. The half-life of Rn is 3.8 days. How many \(\alpha\) particles per second are emitted by decaying Rn nuclei in the room?

Short Answer

Expert verified
What is the number of α particles per second emitted by decaying Rn nuclei in the room, given the half-life of Rn is approximately 3.8 days and there are about 10^7 Rn nuclei in the room? Answer: The isotope of Rn produced by the α decay of radium 226 88 is Rn 222 86. There are 86 protons and 136 neutrons present in its nucleus. The number of α particles per second emitted by decaying Rn nuclei in the room is approximately \(R = \frac{ln(2)}{3.8(24)(3600) \, \text{s}} \times 10^7\) alpha particles per second.

Step by step solution

01

(Step 1: Analyze the decay process)

In an \(\alpha\) decay process, the nucleus loses two protons and two neutrons to emit an \(\alpha\) particle (a helium nucleus). We start by analyzing the decay of radium 226 88 to determine the isotope of Rn produced.
02

(Step 2: Determine the isotope of Rn produced)

Since radium 226 88 loses two protons in the decay, the atomic number of the Rn produced is 88 - 2 = 86. Similarly, the decay of radium 226 88 loses two neutrons, so the mass number of Rn produced is 226 - 4 = 222. Therefore, the isotope of Rn produced in this decay is Rn 222 86.
03

(Step 3: Calculate the number of neutrons and protons)

Since the atomic number Z corresponds to the number of protons and the mass number A corresponds to the sum of protons and neutrons, the number of protons in the Rn 222 86 nucleus is 86, and the number of neutrons is 222 - 86 = 136.
04

(Step 4: Calculate the decay constant, \(\lambda\) )

The decay constant, \(\lambda\) , can be found using the half-life formula: \(\lambda = \frac{ln(2)}{T_{1/2}}\), where \(T_{1/2}\) is the half-life of Rn (3.8 days). So, \(\lambda = \frac{ln(2)}{3.8 \, \text{days}} = \frac{ln(2)}{3.8(24)(3600) \, \text{s}}\)
05

(Step 5: Calculate the decay rate of Rn)

The decay rate is given by the product of the decay constant and the number of Rn nuclei (N) in the room: \(R = \lambda N\). We know there are about \(10^7\) Rn nuclei in the room, so we can calculate the decay rate: \(R = \frac{ln(2)}{3.8(24)(3600) \, \text{s}} \times 10^7\).
06

(Step 6: Calculate the number of \(\alpha\) particles per second)

Since each Rn nucleus decays by emitting an \(\alpha\) particle, the number of \(\alpha\) particles emitted per second is equal to the decay rate of Rn in the room: \(R = \frac{ln(2)}{3.8(24)(3600) \, \text{s}} \times 10^7\) alpha particles per second.

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