Question

What nucleus is obtained in each process? (a) \(\frac{234}{94}\) Pu decays by \(\alpha\) emission. \(\text { (b) } \begin{array}{l}248 \\\97\end{array}\)Bk decays by \(\beta^{-}\) emission. \(\text { (c) } \begin{array}{r}196 \\\82\end{array}\) Pb goes through two successive EC processes.

Short answer

(a) The final nucleus after alpha decay of Pu is Thorium (Th), (b) The final nucleus after beta decay of Bk is Californium (Cf), (c) The final nucleus after two successive electron captures of Pb is Mercury (Hg).

Step-by-step solution

Alpha Decay of Pu

In alpha decay, the atomic number decreases by 2 and the mass number decreases by 4. So, \( \frac{234}{94}\) Pu decays by alpha emission to \( \frac{230}{92}\). Therefore, the resultant nucleus is Thorium (Th). This finding is based on the Periodic Table which shows that the element with an atomic number 92 is Thorium (Th).

Beta Decay of Bk

In beta decay, a neutron in the atomic nucleus is converted into a proton which increases the atomic number by 1 (mass number stays the same). So, \( \frac{248}{97}\) Bk decays by beta emission to \( \frac{248}{98}\). Therefore, the resultant nucleus is Californium (Cf), as per the Periodic Table. The element with an atomic number 98 is Californium (Cf).

Electron Capture of Pb

In electron capture (EC), a proton in the nucleus is converted into a neutron which decreases the atomic number by 1 (mass number stays the same). Since Pb goes through two successive EC processes, the atomic number will decrease by 2 in total. So \( \frac{196}{82}\) Pb after two successive EC reactions becomes \( \frac{196}{80}\). Hence, the resultant nucleus is Mercury (Hg), following the Periodic Table. The element with an atomic number 80 is Mercury (Hg).

Key concepts

Alpha Decay

Alpha decay is a common type of nuclear decay where an unstable nucleus emits an alpha particle, consisting of 2 protons and 2 neutrons. This process results in the formation of a new element. When an alpha particle is emitted:

• The atomic number of the parent nucleus decreases by 2.
• The mass number decreases by 4.

For example, in the case of the original exercise, the isotope of plutonium (\( \frac{234}{94} \) Pu) undergoes alpha decay. The atomic number goes from 94 to 92, and the mass number from 234 to 230. Using the periodic table, we can see the element with atomic number 92 is Thorium (Th). Thus, \( \frac{234}{94} \) Pu decays to \( \frac{230}{92} \) Th. This explains the transformation in terms of change in the element.

Beta Decay

In beta decay, a neutron in the nucleus is transformed into a proton and an electron. This electron, known as a beta particle, is ejected from the nucleus:

• The atomic number increases by 1 because of the new proton.
• The overall mass number remains unchanged.

The case of \( \frac{248}{97} \) Bk (Berkelium) in the exercise is a classic example. When \( \beta^- \) decay occurs, the atomic number increases from 97 to 98, while the mass number remains at 248. According to the periodic table, the element with an atomic number of 98 is Californium (Cf). Hence, \( \frac{248}{97} \) Bk decays to \( \frac{248}{98} \) Cf.

Electron Capture

Electron capture (EC) is a process where an inner orbital electron is captured by the nucleus, which increases the number of neutrons:

• A proton is converted into a neutron, decreasing the atomic number by 1.
• The mass number remains constant.

In the scenario presented, \( \frac{196}{82} \) Pb (lead) undergoes two successive electron capture processes. Each capturing of an electron decreases the atomic number by one. Consequently, two successive EC processes will decrease the atomic number from 82 to 80, with the mass number remaining at 196. The resulting element, according to the periodic table, is Mercury (Hg). Therefore, after two EC processes, \( \frac{196}{82} \) Pb becomes \( \frac{196}{80} \) Hg.