Q24.83P - Chemistry: Molecular Nature Of Matter And Change

Question

The ${}^{\mathbf{\text{80}}}\mathbf{\text{Br}}$ nuclide decays either by decay or by electron capture. 

(a) What is the product of each process? 

(b) Which process releases more energy? (Masses of atoms: ${}^{\mathbf{\text{80}}}\mathbf{\text{Br}}$=$\mathbf{\text{79.918528}}$amu; ${}_{}{}^{\mathbf{\text{80}}}\mathbf{\text{Kr}}$=$\mathbf{\text{79.916380}}$amu; ${}_{}{}^{\mathbf{\text{80}}}\mathbf{\text{Se}}\mathbf{=}\mathbf{\text{79.916520}}$amu; neglect the mass of electrons involved because these are atomic, not nuclear, masses.)

Step-by-Step Solution

Answer

a) 

$$\begin{gathered} {}_{\text{35}}^{\text{80}}\text{Br}{\to }_{\text{ - 1}}^{\text{0}}{\text{e + }}_{\text{36}}^{\text{80}}\text{Kr}{ \\ }_{\text{3}}^{\text{80}}{\text{Br + }}_{\text{ - 1}}^{\text{0}}\text{e}{\to }_{\text{34}}^{\text{80}}\text{Se} \end{gathered}$$

b)Beta decay

Step 1: To find the product of each process

(a)

- Atomic number of Br is $35$

- Beta particle: ${}_{\text{ - 1}}{}^{\text{0}}\text{e}$

Beta decay of ${}^{80}\text{Br}$

$$\begin{gathered} {}_{\text{35}}^{\text{80}}\text{Br}{\to }_{\text{ - 1}}^{\text{0}}{\text{e + }}_{\text{Z}}^{\text{A}}\text{X} \\ \text{ Let us calculate the values of A and Z} \end{gathered}$$

$\text{80 = } \text{0 + AA = } \text{8035 = } \text{ - 1 + ZZ = } \text{36}$

Hence The element with atomic number $36$ is krypton, so the product is krypton-$80$ 

${}_{\text{35}}{}^{\text{80}}\text{Br}{\to }_{\text{ - 1}}^{\text{0}}{\text{e + }}_{\text{36}}^{\text{80}}\text{Kr}$

${}_{}{}^{\text{80}}\text{Br}$electron capture

${}_{35}{}^{80}\text{Br}{+}_{-1}^0\text{e}{\to }_{\text{Z}}^{\text{A}}\text{X}$

Let us calculate the values of A and Z

$\text{80 + 0 = } \text{AA = } \text{80}$

$\text{35 + ( - 1) = } \text{ZZ} \text{ = } \text{34}$

Hence The element with atomic number 34 is selenium, so the product is selenium-$80$ 

${}_{35}{}^{80}\text{Br}{+}_{-1}^0\text{e}{\to }_{34}^{80}\text{Se}$

Step 2: To find which process releases more energy

(b)

$\text{E} \text{ = } {\text{mc}}^{\text{2}}$

- Mass of Br atom is $79.918528$ amu

- Mass of  atom is $\text{79.916380 amu}$

- Mass of Se atom is $\text{79.916520 amu}$

In order to calculate whether beta decay or electron capture releases more energy, we need to calculate the change in mass (mass of product - mass of reactant)

- Beta decay

$\text{∆m} \text{ = } {\text{m}}_{\text{Kr}}{\text{ - m}}_{\text{Br}}\text{ = } \text{79.916380amu - 79.918528} \text{amu} \text{ = } \text{ - 0.002148} \text{amu}$

- Electron capture

$∆\text{m} \text{ = } {\text{m}}_{\text{Se}}{\text{ - m}}_{\text{Br}}\text{ = } \text{79.916520amu - 79.918528amu} \text{ = } \text{ - 0.002008} \text{amu}$

Hence, In both reactions the mass is lost, so in both reactions the energy is released. More mass is lost in beta decay reaction, therefore, more energy is released.