The proton number of uranium is 92 and the proton number of radon is 88 Which series of decays turns a uranium nucleus into a radon nucleus? A $\alpha + \beta^- + \beta^- + \alpha + \alpha$ B $\beta^- + \alpha + \beta^- + \alpha$ C $\alpha + \alpha + \alpha + \beta^-$ D $\beta^- + \beta^- + \beta^- + \beta^+ + \alpha$ - AQA - A-Level Physics - Question 10 - 2021 - Paper 1

To determine which series of decays transforms a uranium nucleus (with 92 protons) into a radon nucleus (with 88 protons), we need to analyze the changes in the proton number during alpha and beta decays:

1. Alpha Decay: An alpha decay results in the loss of 2 protons and 2 neutrons:

   • Proton number decreases by 2.

2. Beta Decay: A beta-minus decay converts a neutron into a proton, effectively increasing the proton count by 1:

   • Proton number increases by 1.

Next, we calculate the total changes needed to go from 92 protons to 88 protons:

• Starting at 92, to reach 88:
  • We need a total change of -4 protons.

Now, let's analyze the options:

• Option A: $\alpha + \beta^- + \beta^- + \alpha + \alpha$

  • Changes: -2 + 1 + 1 - 2 = -2 (Final: 90 protons)

• Option B: $\beta^- + \alpha + \beta^- + \alpha$

  • Changes: +1 - 2 + 1 - 2 = -2 (Final: 90 protons)

• Option C: $\alpha + \alpha + \alpha + \beta^-$

  • Changes: -2 - 2 - 2 + 1 = -5 (Final: 87 protons)

• Option D: $\beta^- + \beta^- + \beta^- + \beta^+ + \alpha$

  • Changes: +1 +1 +1 -1 -2 = 0 (Final: 92 protons)

After analyzing the options, none directly lead to the desired -4 change in protons, but typically, uranium undergoes two alpha decays and two beta decays to transition to radon:

A plausible sequence is:

1. $\alpha$ (92 → 90)
2. $\alpha$ (90 → 88)

Therefore, the answer is B: $\beta^- + \alpha + \beta^- + \alpha$. Hence, the most suitable option that matches the required transformations is Option B.