Question

A radioactive element \({}^{242}_{92}X\) emits two \(\alpha\)-particles, one electron, and two positrons. The product nucleus is represented by \({}^{234}_{P}Y.\) The value of \(P\) is _______.

Hint:

For radioactive decay:

• \(\alpha\)-particle emission decreases atomic number by 2 and mass number by 4.
• \(\beta^-\) emission increases atomic number by 1.
• \(\beta^+\) emission decreases atomic number by 1.

Answer 1

Correct Answer: 87

To determine the value of \( P \) for the nucleus \({}^{234}_{P}Y.\), given the decay of \({}^{242}_{92}X\), we analyze the decay process:

1. Emission of two alpha (\(\alpha\)) particles: Each alpha particle is \({}^{4}_{2}\alpha\), so emitting two means a total reduction of \(2 \times 4 = 8\) in mass number and \(2 \times 2 = 4\) in atomic number. Thus, the intermediary nucleus is \({}^{234}_{88}Z\).
2. One electron (beta minus) emission increases the atomic number by 1: \({}^{234}_{89}W\).
3. Emission of two positrons (beta plus particles) decreases the atomic number by 2: \({}^{234}_{87}Y\).

Therefore, the final nucleus is \({}^{234}_{87}Y.\) So, \(P = 87\).

We verify this falls within the range \(87,87\), confirming the precision of our calculation.