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Chapter 15: Problem 12

A particular nuclide is found to have lost 3 neutrons and 1 proton after a decay chain. What combination of \(\alpha\) and \(\beta\) decays could account for this result?

Short Answer

Expert verified
Answer: The combination of decays that can result in the loss of 3 neutrons and 1 proton is 1 alpha decay and 1 beta decay.

Step by step solution

01

Determine the decay chain equation

Let the combination of alpha decays be represented by 'a' and the combination of beta decays be represented by 'b'. Since in an alpha decay, a nucleus loses 2 protons and 2 neutrons, the total loss of protons in 'a' alpha decays is 2a. In a beta decay, a neutron turns into a proton, so the total gain of protons in 'b' beta decays is b. For the entire decay chain, the loss of 1 proton can be represented by the following equation: 2a - b = 1 Similarly, for neutrons, the total loss of neutrons in 'a' alpha decays is 2a, and the total loss of neutrons in 'b' beta decays is b. The total loss of 3 neutrons in the entire decay chain is represented by the following equation: 2a + b = 3
02

Solve the system of equations

Now, we have a system of two linear equations with two variables, a and b: 1) 2a - b = 1 2) 2a + b = 3 We can solve these equations using different methods, but let's use the elimination method. Add equation (1) and equation (2) to eliminate the 'b' variable: 2a - b + 2a + b = 1 + 3 4a = 4 Now, let's solve for 'a': a = 4/4 a = 1 To find the value of 'b', substitute the value of 'a' into either equation, such as equation (1): 2(1) - b = 1 2 - b = 1 b = 1
03

Final result

The combination of alpha (a) and beta (b) decays that can account for the loss of 3 neutrons and 1 proton after a decay chain is: a = 1 (one alpha decay) b = 1 (one beta decay)

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Most popular questions from this chapter

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