Home / CIE AS & A Level Physics : 11.1 Atoms, nuclei and radiation – Exam style question – Paper 1
A nucleus of polonium, \(^{214}_{84}\mathrm{Po}\), decays by emitting an \(\alpha\)-particle to become a nucleus of lead.
The nucleus of lead is also unstable and decays by emitting a \( \beta^{-} \) particle to form a nucleus of bismuth which then decays by emitting a \( \beta^{-} \) particle to produce a nucleus \(X\).
What is the number of neutrons in nucleus \(X\)?
Correct Answer: \( \boxed{\mathrm{A}} \)
After \(\alpha\)-decay:
\(^{214}_{84}\mathrm{Po}\rightarrow{}^{210}_{82}\mathrm{Pb}\).
Each \( \beta^{-} \) decay increases the atomic number by \(1\) while the mass number remains \(210\).
After two \( \beta^{-} \) decays, nucleus \(X\) is \(^{210}_{84}\mathrm{Po}\).
Number of neutrons \(=210-84=126\).
Therefore, the correct answer is (A).
The diagram shows a sequence of radioactive decays involving three \( \alpha \)-particles and one \( \beta^- \) particle.
What is nuclide \( T \)?
Correct Answer: \( \boxed{\mathrm{A}} \)
Starting with \( ^{237}_{93}\mathrm{Np} \):
Each \( \alpha \)-decay decreases the mass number by \( 4 \) and the proton number by \( 2 \).
One \( \beta^- \)-decay increases the proton number by \( 1 \) but leaves the mass number unchanged.
After three \( \alpha \)-decays and one \( \beta^- \)-decay:
Mass number \( =237-12=225 \).
Proton number \( =93-6+1=88 \).
Hence, \( T=^{225}_{88}\mathrm{Ra} \).
Therefore, the correct answer is (A).
A stationary nucleus has nucleon number \(A\).
The nucleus decays by emitting a proton with speed \(v\) to form a new nucleus with speed \(u\). The new nucleus and the proton move away from one another in opposite directions.
Which equation gives \(v\) in terms of \(A\) and \(u\)?
Correct Answer: \( \boxed{\mathrm{B}} \)
Initially, the nucleus is stationary, so the total momentum is zero.
Applying conservation of momentum,
\(m_pv=(A-1)m_pu\).
Cancelling the proton mass \(m_p\) gives
\(v=(A-1)u\).
Therefore, the correct answer is (B).