Question Video: Identifying the Correct Equation for the Beta Decay of Neptunium-239 Chemistry • First Year of Secondary School

Video Transcript

Which of the following equations represents the 𝛽 decay of neptunium-239? (A) 239 93 Np forms 235 95 Am plus 0 −1 e plus 4 2 He. (B) 239 93 Np plus 0 −1 e forms 239 94 Pu. (C) 239 93 Np plus 0 −1 e forms 239 93 Np. (D) 239 93 Np forms 239 94 Pu plus 0 −1 e. Or (E) 239 93 Np forms 239 92 U plus 0 −1 e.

To answer this question, we have to know what’s in the nucleus of neptunium-239 and what happens during 𝛽 decay. Then we can apply our equation for nuclear reactions to determine the 𝑍 and 𝐴 numbers of the products. First, what is in the nucleus of neptunium-239? Neptunium is element 93. So it has a 𝑍 of 93. We are told that it’s neptunium-239. So its mass number is 239. This means that the value for 𝐴 is 239, which is the total number of protons and neutrons in the nucleus. We can then write neptunium-239 as 239 93 Np.

Next, what happens during 𝛽 decay? During 𝛽 decay, a 𝛽 particle is released. A 𝛽 particle is essentially an electron. It has a 𝑍 of minus one and an 𝐴 of zero since it is a negatively charged particle with a very tiny mass in comparison to protons and neutrons. The symbol 0 −1 e can be used. As the 𝛽 particle is released, it should be on the product side of the equation, not the reactant side. Thus, we can rule out options (B) and (C) as being the correct answer to this question.

Now we can apply our equation for nuclear reactions. For 𝑍, we get this equation, where 𝑍 products equals 𝑍 reactants. Thus, 𝑍 for the new isotope plus 𝑍 for the 𝛽 particle equals 𝑍 for neptunium-239. We know the value of 𝑍 for the 𝛽 particle and for neptunium-239. So we need to rearrange the equation to make 𝑍 for the new isotope the subject. If we rearrange the equation and substitute the values, then we find that the value of 𝑍 for the new isotope is 94. Of the answer choices that we haven’t already ruled out, the only one that has a value of 94 for 𝑍 for the new isotope is option (D). So option (D) must be the correct answer to this question, where the new isotope is plutonium.

But for completeness, let’s work out the value of 𝐴, the number of particles in the nucleus. If we substitute the known values of 𝐴, we get a value of 239. Thus, our product is plutonium-239. Therefore, the equation that represents the 𝛽 decay of neptunium-239 is (D) 239 93 Np forms 239 94 Pu plus 0 −1 e.