Problem

Write the nuclear equations for the beta decay of the following isotopes:
(a) ${}_{82}\mathrm{Pb}^{210}$
(b) ${}_{83}\mathrm{Bi}^{210}$
(c) ${}_{90}\mathrm{Th}^{234}$
(d) ${}_{93}\mathrm{Np}^{239}$

Data

• Type of decay: Beta decay ($\beta^-$-decay)
• General formula:

$${}_{Z}^{A} X \rightarrow {}_{Z+1}^{A} Y + {}_{-1} \beta + \bar{v} + Q$$

Where:

• $Z$: Atomic number of the parent nucleus
• $A$: Mass number (remains unchanged)
• $Y$: Daughter nucleus with atomic number $Z+1$
• $\beta^-$: Emitted beta particle (electron)
• $\bar{v}$: Emitted antineutrino
• $Q$: Released energy

Prerequisite Concepts

1. Beta Decay ($\beta^-$-decay):

   • Occurs when a neutron in the nucleus converts into a proton, emitting a beta particle ($\beta^-$) and an antineutrino ($\bar{v}$).
   • The atomic number $Z$ increases by 1, while the mass number $A$ remains unchanged.

2. Nuclear Equation Format:
   The parent nucleus ${}_{Z}^{A} X$ transforms into the daughter nucleus ${}_{Z+1}^{A} Y$, emitting:

   • A beta particle ($\beta^-$)
   • An antineutrino ($\bar{v}$)

Solution

(a) Beta decay of ${}_{82}\mathrm{Pb}^{210}$:

• Parent nucleus: ${}_{82}\mathrm{Pb}^{210}$
• Daughter nucleus: ${}_{83}\mathrm{Bi}^{210}$

Nuclear equation:

$${}_{82}\mathrm{Pb}^{210} \rightarrow {}_{83}\mathrm{Bi}^{210} + {}_{-1}\beta + \bar{v} + Q$$

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(b) Beta decay of ${}_{83}\mathrm{Bi}^{210}$:

• Parent nucleus: ${}_{83}\mathrm{Bi}^{210}$
• Daughter nucleus: ${}_{84}\mathrm{Po}^{210}$

Nuclear equation:

$${}_{83}\mathrm{Bi}^{210} \rightarrow {}_{84}\mathrm{Po}^{210} + {}_{-1}\beta + \bar{v} + Q$$

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(c) Beta decay of ${}_{90}\mathrm{Th}^{234}$:

• Parent nucleus: ${}_{90}\mathrm{Th}^{234}$
• Daughter nucleus: ${}_{91}\mathrm{Pa}^{234}$

Nuclear equation:

$${}_{90}\mathrm{Th}^{234} \rightarrow {}_{91}\mathrm{Pa}^{234} + {}_{-1}\beta + \bar{v} + Q$$

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(d) Beta decay of ${}_{93}\mathrm{Np}^{239}$:

• Parent nucleus: ${}_{93}\mathrm{Np}^{239}$
• Daughter nucleus: ${}_{94}\mathrm{Pu}^{239}$

Nuclear equation:

$${}_{93}\mathrm{Np}^{239} \rightarrow {}_{94}\mathrm{Pu}^{239} + {}_{-1}\beta + \bar{v} + Q$$

Answer

1. ${}_{82}\mathrm{Pb}^{210} \rightarrow {}_{83}\mathrm{Bi}^{210} + {}_{-1}\beta + \bar{v} + Q$
2. ${}_{83}\mathrm{Bi}^{210} \rightarrow {}_{84}\mathrm{Po}^{210} + {}_{-1}\beta + \bar{v} + Q$
3. ${}_{90}\mathrm{Th}^{234} \rightarrow {}_{91}\mathrm{Pa}^{234} + {}_{-1}\beta + \bar{v} + Q$
4. ${}_{93}\mathrm{Np}^{239} \rightarrow {}_{94}\mathrm{Pu}^{239} + {}_{-1}\beta + \bar{v} + Q$