CIE iGCSE Co-Ordinated Science P5.2.3 Radioactive decay Exam Style Questions Paper 2
▶️ Answer/Explanation
In \(\beta^-\) decay, a neutron changes into a proton, so the proton number increases by 1 (from 6 to 7) while the nucleon number remains unchanged (14). So \(A = 14\) and \(Z = 7\).
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
In the decay equation, the nucleon number (90) stays the same, but the proton number increases from 38 to 39. This means a neutron in the nucleus has converted into a proton, emitting a beta particle (electron) in the process. The emitted particle must therefore be a beta particle. This is characteristic of beta decay where a neutron changes into a proton, increasing the atomic number by 1 with no change in mass number.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
In alpha decay, the nucleus loses an alpha particle (\(^{4}_{2}\text{He}\)), so nucleon number decreases by 4 and proton number decreases by 2. From \(^{238}_{92}\text{U}\), \(x = 238 – 4 = 234\) and \(y = 4\). Thorium is \(^{234}_{90}\text{Th}\).
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
An alpha particle is \(_2^4\textrm{He}\), so emitting one gives \(_{92}^{238}\textrm{U} \rightarrow \, _{90}^{234}\textrm{Th} + \, _2^4\textrm{He}\).
Mass number: \(238-4=234\); proton number: \(92-2=90\).
✅ Answer: (A)
Mass number: \(238-4=234\); proton number: \(92-2=90\).
✅ Answer: (A)
Question
A uranium nucleus decays into a thorium nucleus. The thorium nucleus then decays into a protactinium nucleus.
$_{238}^{92}\textrm{U}\to _{234}^{90}\textrm{Th}\to _{234}^{91}\textrm{Pa}$
Which emissions take place during the decays?
(A) an alpha-particle followed by a beta-particle
(B) an alpha-particle followed by another alpha-particle
(C) a beta-particle followed by an alpha-particle
(D) a beta-particle followed by another beta-particle
▶️Answer/Explanation
Ans: A
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