Video: Solving Nuclear Equations Involving Beta Decay

The nuclear equation, shows how argon decays to potassium via beta decay. What is the value of π‘š in the equation?

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Video Transcript

The nuclear equation, shown here, shows how argon decays to potassium via beta decay. What is the value of π‘š in the equation?

Taking a look at this equation, we see that, indeed, the element argon is decaying into potassium plus a beta particle. Looking at the numbers corresponding to each of these atomic symbols, we see that argon has 18 protons and three nine protons plus neutrons. Going over to potassium, potassium has π‘š protons and 39 protons plus neutrons. Meanwhile, our beta particle, symbolised over here, is an electron. Therefore, it has a charge of negative one. And it has zero protons plus neutrons.

When it comes to the charge on an electron, such as the charge of a beta particle and the charge of a proton, we know these charges are equal and opposite. That is, they have the same magnitude, but the opposite sign. The charge of a proton is positive while the charge of an electron is negative.

When our question asks what the value of π‘š is in the equation, where π‘š is the number of protons and potassium, we can solve for this by realising that, across this nuclear equation, electric charge is conserved. The total electric charge in this reaction before the decay occurs must equal the total electric charge after the decay occurs.

If we let positive one represent the charge of a proton and negative one represent the charge of an electron, then we can say that, initially, in this reaction, we have 18 times positive one electric charge, 18 protons. Then, after the decay event, we have π‘š, an unknown number we want to solve for, plus one times negative one because there’s one electron in the beta particle. We can simplify this. So it reads 18 yields π‘š minus one.

Since π‘š minus one is equal to 18, that must mean that π‘š is equal to 19. That’s the number of protons in potassium. And it’s the value of π‘š in this equation.

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