10. examine the possibility of making gold atoms by alpha or beta decay as shown in these two illustrations…

10. examine the possibility of making gold atoms by alpha or beta decay as shown in these two illustrations. place the appropriate starting isotope in each equation. either write again here with your mouse/stylus, or complete the worksheet on your coursepack or journal and upload photos. :0) alpha decay: ₇₉¹⁹⁷au + α beta decay: ₇₉¹⁹⁷au + β

10. examine the possibility of making gold atoms by alpha or beta decay as shown in these two illustrations. place the appropriate starting isotope in each equation. either write again here with your mouse/stylus, or complete the worksheet on your coursepack or journal and upload photos. :0) alpha decay: ₇₉¹⁹⁷au + α beta decay: ₇₉¹⁹⁷au + β

Answer

Explanation:

Step1: Recall beta - decay concept

In beta - decay, a neutron in the nucleus is converted into a proton, an electron (beta - particle), and an antineutrino. The atomic number increases by 1 while the mass number remains the same. For gold - 197 ($^{197}{79}Au$), the beta - decay equation is $^{197}{79}Au\rightarrow^{197}_{80}Hg + _{- 1}^0e+\bar{\nu}_e$.

Step2: Recall alpha - decay concept

In alpha - decay, an alpha - particle ($^{4}{2}He$) is emitted from the nucleus. The atomic number decreases by 2 and the mass number decreases by 4. For gold - 197 ($^{197}{79}Au$), if it were to undergo alpha - decay, the equation would be $^{197}{79}Au\rightarrow^{193}{77}Ir+^{4}_{2}He$. But gold - 197 is a stable isotope in nature, so it does not undergo spontaneous alpha or beta decay under normal conditions.

Answer:

Beta - decay: $^{197}{79}Au\rightarrow^{197}{80}Hg + {- 1}^0e+\bar{\nu}e$ Alpha - decay: $^{197}{79}Au\rightarrow^{193}{77}Ir+^{4}_{2}He$ (but gold - 197 is stable and does not decay spontaneously in these ways in nature)