The radioactive isotope \(^{242} \mathrm{Cm}\) decays by a series of \(\alpha\) -particle and \(\beta\) -particle productions, taking \(^{242} \mathrm{Cm}\) through many transformations to end up as \(^{206} \mathrm{P} \mathrm{b}\) . In the complete decay series, how many \(\alpha\) and \(\beta\) particles are produced?

In this exercise, we are given that the initial isotope is \(^{242} \mathrm{Cm}\), and the final isotope is \(^{206} \mathrm{Pb}\). The atomic number of the initial isotope (Cm) is 96, and the atomic number of the final isotope (Pb) is 82.

We need to find the change in mass and the change in atomic number during the complete decay process. Change in mass = Initial mass - Final mass = \(242 - 206 = 36\) Change in atomic number = Initial atomic number - Final atomic number = \(96 - 82 = 14\)

Let x be the number of alpha particles and y be the number of beta particles produced during the decay process. Since an alpha particle decreases the atomic mass by 4, the total decrease in atomic mass due to alpha particle production is \(4x\). As beta particles do not cause a change in mass, no change is due to beta particles in this case. Similarly, since an alpha particle decreases the atomic number by 2, the total decrease in atomic number due to alpha particle production is \(2x\). A beta particle increases the atomic number by 1, so the total increase in atomic number due to beta particle production is \(y\). Now we can write two equations: 1. Change in mass = \(4x\) 2. Change in atomic number = \(2x + y\) Using the values calculated in Step 2, we have: 1. 36 = \(4x\) 2. 14 = \(2x + y\)

First, solve for x from the first equation: 36 = \(4x \Rightarrow x = \frac{36}{4} = 9\) Now substitute this value of x in the second equation: 14 = \(2(9) + y \Rightarrow 14 = 18 + y \Rightarrow y = -4\) Since we have a negative value for y, we made an error in understanding the decay process. Beta decay increases the atomic number (not decreases). Therefore, the correct second equation should be: 2. Change in atomic number = \(2x - y\) Now, use the new second equation and the value of x: 14 = \(2(9) - y \Rightarrow 14 = 18 - y \Rightarrow y = 4\)

We found that during the decay process of \(^{242} \mathrm{Cm}\) to \(^{206} \mathrm{Pb}\), the number of alpha particles produced is 9 and the number of beta particles produced is 4.