The radioactive isotope \({ }^{247} \mathrm{Bk}\) decays by a series of \(\alpha\) -particle and \(\beta\) -particle productions, taking \({ }^{247} \mathrm{Bk}\) through many transformations to end up as \({ }^{207} \mathrm{~Pb}\). In the complete decay series, how many \(\alpha\) particles and \(\beta\) particles are produced?

Since we know the initial and final elements in the series, we can determine the change in mass number. The mass number decreases from 247 (in \({ }^{247} \mathrm{Bk}\)) to 207 (in \({ }^{207} \mathrm{~Pb}\)), a total decrease of: Change in mass number = Initial mass number - Final mass number \( \Delta A = 247 - 207 \) #Step 2: Calculate the number of alpha particles emitted#

The decrease in mass number should be equal to the total number of alpha particles multiplied by 4, as each alpha particle reduces the mass number by 4. So we can now calculate the number of alpha particles emitted during the decay process: Number of alpha particles = Change in mass number / 4 \( n_\alpha = \Delta A / 4 \) #Step 3: Calculate the change in the atomic number# Since we are not given the atomic numbers of Bk (Berkelium) and Pb (Lead), we need to find that information. We can look it up in a periodic table. Berkelium has an atomic number of 97, while lead has an atomic number of 82. Therefore, the change in atomic number is: Change in atomic number = Initial atomic number - Final atomic number \( \Delta Z = 97 - 82 \) #Step 4: Calculate the number of beta particles emitted#

Now, we know that each alpha particle decreases the atomic number by 2, while each beta particle increases it by 1. Using the change in atomic number and the number of alpha particles already calculated, we can calculate the number of beta particles: Number of beta particles = Change in atomic number + 2 * Number of alpha particles \( n_\beta = \Delta Z + 2 \times n_\alpha \) #Solution# #Step 1: Analyze the change in atomic and mass numbers#

Change in mass number = Initial mass number - Final mass number \( \Delta A = 247 - 207 = 40 \) #Step 2: Calculate the number of alpha particles emitted#

Number of alpha particles = Change in mass number / 4 \( n_\alpha = \Delta A / 4 = 40 / 4 = 10 \) #Step 3: Calculate the change in the atomic number# Change in atomic number = Initial atomic number - Final atomic number \( \Delta Z = 97 - 82 = 15 \) #Step 4: Calculate the number of beta particles emitted#

Number of beta particles = Change in atomic number + 2 * Number of alpha particles \( n_\beta = \Delta Z + 2 \times n_\alpha = 15 + 2 \times 10 = 15 + 20 = 35 \) In conclusion, during the complete decay series of \({ }^{247} \mathrm{Bk}\) to \({ }^{207} \mathrm{~Pb}\), 10 \(\alpha\) particles and 35 \(\beta\) particles are produced.