The radioactive nuclei 86Ra228 decays by a series of emissions of three alpha particles and one beta particle. The nuclei X formed is:
Correct answer: B. 81X216
- A. 84X220
- B. 81X216
- C. 86X222
- D. 88X215
Explanation
The following is the solution: There are a total of 3 alpha emissions each alpha molecule is α42 which means a decrease in mass number by 4 and atomic number by 2 with each alpha particle release. (W, X, Y, and Z denote the released atoms). Ra22886---> W22484 ---> Y22082 ---> Z21680 There is 1 beta emission each beta molecule is β0-1 which means no change in mass number and the atomic number increases by 1. Z21680 ---> X21681
Last updated
About Nuclear Physics
Atomic nuclei contain protons and neutrons, with nuclear size, mass defect and binding energy describing their stability. The work covers alpha, beta and gamma decay, decay equations, half-life and decay rate, along with radiation safety and medical applications such as imaging, radiotherapy and tracers.
Practise Nuclear Physics
1,254 free Nuclear Physics MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Physics questions like this
Physics is on 13 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.