(b) A radioactive isotope of Americium (Am —241) decays into a nucleus of Neptunium (Np — 237) and an alpha (\(\alpha\)) particle as shown in the nuclear equation below. \(^{241}_{95}Am\) \(\to\) \(^{237}_{c} + ^b_a \alpha\)
(i) State the number of neutrons in the nucleus of Americium — 241.
(ii) Determine the values of a,b and c.
(ii) State two properties of gamma rays that make them suitable for sterilizing medical equipment.
(d) A sample of radioactive substance was found to be left with h of its initial count rate after 110 years. Calculate its decay constant.
(a) Nucleon number (mass number) is the total number of protons and neutrons contained in the nucleus of an atom.
(b) The decay is \({}^{241}_{95}\text{Am} \to {}^{237}_{c}\text{Np} + {}^{b}_{a}\alpha\).
(i) Number of neutrons in \(\text{Am}\text{-}241 = \text{mass number} - \text{proton number} = 241 - 95 = 146\).
(ii) An alpha particle is a helium nucleus, \({}^{4}_{2}\text{He}\), so \(a = 2\) and \(b = 4\).
Conserving proton (atomic) number: \(95 = c + 2 \Rightarrow c = 93\).
Check with mass number: \(241 = 237 + 4\), which balances. Hence \(a = 2,\ b = 4,\ c = 93\).
(c)(i) Gamma rays are not deflected by an electric or magnetic field because they are electromagnetic waves and carry no electric charge; only charged particles experience a deflecting force in such fields.
(c)(ii) Two properties that make gamma rays suitable for sterilising medical equipment:
- They have very high penetrating power, so they can reach and destroy micro-organisms even inside sealed or packaged equipment.
- They are strongly lethal to bacteria and other micro-organisms, and they do so without leaving any residue or making the equipment radioactive.
(d) The sample is left with \(\dfrac{1}{32}\) of its initial count rate after \(110\) years. Writing the fraction as a power of one half:
\[\frac{N}{N_0} = \frac{1}{32} = \left(\frac{1}{2}\right)^{5},\]
so \(n = 5\) half-lives have elapsed. The half-life is
\[t_{1/2} = \frac{\text{total time}}{n} = \frac{110}{5} = 22\ \text{years}.\]
The decay constant is
\[\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{22} = 3.15\times10^{-2}\ \text{year}^{-1}\ (\approx 3.2\times10^{-2}\ \text{year}^{-1}).\]
Examination note: the neutron count is always mass number minus proton number, so use \(241 - 95 = 146\), not \(241 - 90\). For the half-life step, express the remaining fraction as \(\left(\tfrac{1}{2}\right)^{n}\) to read off the number of half-lives before using \(\lambda = \dfrac{\ln 2}{t_{1/2}}\).
