19. A radioactive nucleus decays to yield a bismuth-211 nucleus and an alpha particle. What was the identity of the original nucleus? Show the nuclear equation that leads you to this answer.
It's hard to do subscripts and superscripts here so here is what I will do.
A is the atomic number, M is the mass number, X is the element. so
AX^M ==> 2He^4 + 83Bi^211
Now you want the subscripts to add up on both sides as well as the superscripts to add up on both sides.
A = 83 + 2 = 85 and you look on the periodic table to find the identity of X
M = 211 + 4 = ?
Now fill in the AX^N we started with.
85At^215 = 2He^4 + 83Bi^211
To find the identity of the original nucleus, we need to understand the process of radioactive decay and the properties of the involved particles.
In this scenario, we know that the original nucleus decays to yield a bismuth-211 nucleus and an alpha particle (α). An alpha particle consists of two protons and two neutrons, which means it has a mass number of 4 and an atomic number of 2 (since the atomic number represents the number of protons in an element).
Let's represent the original nucleus as X:
X → bismuth-211 + alpha particle
Now, we can use the conservation of mass number and atomic number to determine the identity of the original nucleus.
The mass number is the sum of protons and neutrons, which remains constant throughout decay reactions. Therefore, we can write the equation as:
A(X) → 211(Bi) + 4(α)
The atomic number represents the number of protons, and it also remains constant throughout the decay process. By comparing the atomic numbers on both sides of the equation, we can determine the atomic number of the original nucleus.
The atomic number of bismuth is 83, and the atomic number of an alpha particle is 2. Therefore, the atomic number of the original nucleus can be calculated by subtracting the atomic numbers of bismuth (83) and the alpha particle (2) from the total atomic number of the original nucleus.
Let's represent the atomic number of the original nucleus as Z:
Z(X) → 83(Bi) + 2(α)
Z - 2 = 83 - 2
Z - 2 = 81
Z = 81 + 2
Z = 83
Hence, the identity of the original nucleus is lead-83 (Pb-83).
The nuclear equation representing the decay is:
Pb-83 → 211(Bi) + 4(α)