Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \({ }_{1}^{3} \mathrm{H}(\beta)\) b. \({ }_{3}^{8} \mathrm{Li}(\beta\) followed by \(\alpha\) ) c. \({ }_{4}^{7}\) Be (electron capture) d. \({ }_{5}^{8}\) B (positron)

Short Answer

Expert verified
a. \({ }_{1}^{3} \mathrm{H} \rightarrow { }_{2}^{3} \mathrm{He} + e^{-} + \overline{\nu}\) b. \({ }_{3}^{8} \mathrm{Li} \rightarrow { }_{4}^{8} \mathrm{Be} + e^{-} + \overline{\nu}\); \({ }_{4}^{8} \mathrm{Be} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{2}^{4} \mathrm{He}\) c. \({ }_{4}^{7} \mathrm{Be} + e^{-} \rightarrow { }_{3}^{7} \mathrm{Li} + \nu\) d. \({ }_{5}^{8} \mathrm{B} \rightarrow { }_{4}^{8} \mathrm{Be} + e^{+} + \nu\)

Step by step solution

01

Write the decay process

The nuclide H-3 undergoes beta decay, in which a neutron is transformed into a proton with the emission of an electron (β) and an antineutrino (ν). The equation for the decay process is as follows: \[ { }_{1}^{3} \mathrm{H} \rightarrow P + e^{-} + \overline{\nu} \]
02

Write the balanced decay equation

To balance the decay equation, we need to ensure that the atomic number and mass number are conserved. The decay product P is a helium isotope with one proton. The equation becomes: \[ { }_{1}^{3} \mathrm{H} \rightarrow { }_{2}^{3} \mathrm{He} + e^{-} + \overline{\nu} \] #b. Li-8 (β followed by α) decay#
03

Write β-decay process

The nuclide Li-8 first undergoes β-decay, in which a neutron is transformed into a proton with the emission of an electron and an antineutrino. The equation for the β-decay process is as follows: \[ { }_{3}^{8} \mathrm{Li} \rightarrow P + e^{-} + \overline{\nu} \]
04

Write balanced β-decay equation

The atomic number and mass number should be conserved in the decay process. The decay product P is a beryllium isotope. Therefore, the balanced β-decay equation is: \[ { }_{3}^{8} \mathrm{Li} \rightarrow { }_{4}^{8} \mathrm{Be} + e^{-} + \overline{\nu} \]
05

Write α-decay process

The decay product, Be-8, then undergoes α-decay. In α-decay, the nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons (a helium-4 nucleus). The equation for the α-decay process is as follows: \[ { }_{4}^{8} \mathrm{Be} \rightarrow P + { }_{2}^{4} \mathrm{He} \]
06

Write balanced α-decay equation

Applying conservation of atomic number and mass number, the decay product P is a helium-4 isotope. The balanced α-decay equation is: \[ { }_{4}^{8} \mathrm{Be} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{2}^{4} \mathrm{He} \] #c. Be-7 (electron capture) decay#
07

Write the decay process

In electron capture, a nucleus absorbs an electron from the innermost shell, and a proton is transformed into a neutron with the emission of a neutrino (ν). The equation for this decay process is as follows: \[ { }_{4}^{7} \mathrm{Be} + e^{-} \rightarrow P + \nu \]
08

Write the balanced decay equation

To balance the decay equation, we need to ensure that the atomic number and mass number are conserved. The decay product P is a lithium isotope. The equation becomes: \[ { }_{4}^{7} \mathrm{Be} + e^{-} \rightarrow { }_{3}^{7} \mathrm{Li} + \nu \] #d. B-8 (positron) decay#
09

Write the decay process

In positron emission, a proton is transformed into a neutron with the emission of a positron (β⁺) and a neutrino. The decay equation for this process is as follows: \[ { }_{5}^{8} \mathrm{B} \rightarrow P + e^{+} + \nu \]
10

Write the balanced decay equation

To balance the decay equation, we need to ensure that the atomic number and mass number are conserved. The decay product P is a beryllium isotope. The equation becomes: \[ { }_{5}^{8} \mathrm{B} \rightarrow { }_{4}^{8} \mathrm{Be} + e^{+} + \nu \]

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