Lead Isotope Mass: Thorium Decay & Nuclear Reaction Analysis

Hey there, chemistry enthusiasts! Let's dive into a fascinating nuclear chemistry problem. We're going to figure out the relative atomic mass of a lead isotope that forms when thorium undergoes radioactive decay. This kind of problem is super interesting because it combines our knowledge of nuclear reactions, isotopes, and mass calculations. Ready to break it down? Let's get started!

Understanding the Problem: Thorium Decay and Nuclear Reactions

Alright, guys, here’s the scenario: We're starting with 11.25 mg of thorium (specifically, $^{225}_{90}Th$). This thorium undergoes a nuclear reaction, which is basically a fancy way of saying it breaks down into other elements. The reaction spits out a bunch of particles, and we’re told that it releases a whopping $12.04 imes 10^{19}$ electrons. The specific nuclear reaction we're dealing with is: $^{225}_{90}Th ightarrow Pb + x ^4_2 He + y ^0_{-1} e + 5 ^1_0 n$. Our main goal? To determine the relative atomic mass of the lead (Pb) isotope produced in this reaction. This means we need to find out the specific mass number of the lead isotope. To do this, we'll need to use some core concepts from nuclear chemistry, including understanding how radioactive decay works, the types of particles involved (like alpha particles and electrons), and how to balance nuclear equations. We'll also need to use the given information (mass of thorium, number of electrons released) to work backward and figure out the unknowns in the reaction.

So, what's our game plan? First off, we'll need to figure out how many alpha particles ($^4_2 He$) and electrons ($^0_{-1} e$) are released during the reaction. Then, using the law of conservation of mass and charge, we can identify the lead isotope. Finally, knowing the number of neutrons released, we will find the relative atomic mass of the lead. Does it sound like a lot? Trust me, with step-by-step logic, we’ll make it easy! Let's go!

This kind of problem underscores how crucial it is to understand nuclear transformations. Nuclear reactions, unlike ordinary chemical reactions, involve changes within the nucleus of an atom, leading to the formation of new elements and the release of substantial energy. These reactions are governed by conservation laws. These conservation laws are the cornerstone of solving such nuclear problems, allowing us to accurately predict the products of a nuclear reaction. So, are you ready to solve this kind of problems? I know you are! Let's get cracking!

Step-by-Step Solution: Unraveling the Nuclear Reaction

Okay, folks, let's roll up our sleeves and solve this. First things first: We know that the reaction starts with thorium ($^{225}_{90}Th$) and produces lead (Pb), alpha particles ($^4_2 He$), electrons ($^0_{-1} e$), and neutrons ($^1_0 n$). The balanced equation will help us immensely in identifying the lead isotope. We're given the equation: $^{225}_{90}Th ightarrow Pb + x ^4_2 He + y ^0_{-1} e + 5 ^1_0 n$. Our mission is to figure out the values of x and y, which represent the number of alpha particles and electrons, respectively. Let's do this step by step.

1. Calculate the number of Thorium atoms: We're given the mass of thorium (11.25 mg) and we know the molar mass of $^{225}_{90}Th$ (approximately 225 g/mol). Let's convert the mass of thorium into grams and then into moles. Then, use Avogadro's number ($6.022 imes 10^{23}$ atoms/mol) to find the total number of thorium atoms.

• Convert mass to grams: 11.25 mg = $11.25 imes 10^{-3}$ g
• Calculate moles: $(11.25 imes 10^{-3} g) / 225 g/mol = 5 imes 10^{-5}$ mol
• Calculate the number of thorium atoms: $(5 imes 10^{-5} mol) imes (6.022 imes 10^{23} atoms/mol) = 3.011 imes 10^{19}$ atoms.

2. Find the Number of Alpha Particles (x): We know that each thorium atom decays and releases some number of alpha particles. Also, each alpha particle carries 2 positive charges and the total number of electrons released during the reaction. The problem tells us that $12.04 imes 10^{19}$ electrons are emitted. Alpha particles are positively charged, while electrons are negatively charged. Let's consider how these charges balance out in the reaction:

• Each alpha particle has a charge of +2.
• Each electron has a charge of -1.

Now, let's consider the number of alpha particles as x. The total charge contributed by alpha particles is $2x$. The total charge contributed by the electrons is $-y$. The thorium nucleus decays, and the charge must be conserved, so we have the relation:

• $2x - y = 0$

But we already know that during the process, $3.011 imes 10^{19}$ thorium atoms decay, resulting in $12.04 imes 10^{19}$ electrons. This means that for each decaying thorium atom, an average number of electrons are released. Since each electron has a charge of -1 and each alpha particle has a charge of +2. So, we need to balance this charge. The total number of electrons is given. From these two values, we can calculate how many alpha particles are released during the decay. From this logic, we deduce that the number of alpha particles $x = (12.04 imes 10^{19} / 2) / 3.011 imes 10^{19}$. So, the number of alpha particles emitted per thorium atom, we get $x = 2$.

3. Find the Number of Electrons (y):

Now that we know x (the number of alpha particles), we can figure out y (the number of electrons) by considering the total number of electrons released in the decay, which is given in the problem as $12.04 imes 10^{19}$. This means that each thorium atom emits a lot of electrons. So, now we can solve for y by considering that the number of emitted electrons is given as $12.04 imes 10^{19}$ electrons. Thus $y = 12.04 imes 10^{19}$.

4. Balance the Nuclear Equation:

Now, plug in the values of x and y into the balanced equation: ^{225}_{90}Th ightarrow Pb + 2 ^4_2 He + 12.04 imes 10^{19} ^0_{-1} e + 5 ^1_0 n.

Now let's balance the equation for mass numbers (superscripts) and atomic numbers (subscripts) to determine the lead isotope. The sum of mass numbers on the left should equal the sum on the right:

• 225 = A + (2 * 4) + (5 * 1)

So A = 225 - 8 - 5 = 212. Thus, the mass number for lead is 212.

Now, let's balance the atomic numbers:

• 90 = Z + (2 * 2) - 12.04*10^19 + 5 * 0

So Z = 90 - 4 + 12.04 * 10^19 = 82. Lead's atomic number is 82. So, we know that the lead isotope is $^{212}_{82}Pb$.

Determining the Relative Atomic Mass of the Lead Isotope

Alright! Now that we know the lead isotope is $^{212}_{82}Pb$, we know the relative atomic mass is simply the mass number of the isotope. This is because the mass number represents the total number of protons and neutrons in the nucleus, which accounts for almost all of the atom's mass.

Therefore, the relative atomic mass of the lead isotope produced in the reaction is 212.

This means that the relative atomic mass of the lead isotope is approximately equal to its mass number, 212 atomic mass units (amu). This value represents the total number of protons and neutrons in the lead nucleus, which is the major contributor to the atom's mass.

Conclusion: Mastering Nuclear Decay Problems

There you have it! We've successfully calculated the relative atomic mass of the lead isotope produced during the thorium decay. This problem demonstrates the practical application of nuclear reaction principles, including conservation laws, and the importance of understanding the relationship between the number of particles. By carefully tracking the particles involved, balancing the nuclear equation, and using basic arithmetic, we were able to arrive at the solution.

This journey highlights the power of nuclear chemistry to predict and explain the behavior of radioactive isotopes. Keep practicing, and you'll become a pro at solving these types of problems! If you're interested in nuclear chemistry or simply enjoy solving challenging problems, this is definitely a fascinating area to explore. Hope you found this walkthrough helpful and enjoyable. Keep exploring, and you'll master these problems in no time. See you later, chemistry wizards!