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Radioactive decay, half-life, decay constant, activity, and simple cubic lattice
Question:
Polonium-209 is a radioactive metal discovered by Marie Curie, a Polish chemist, in 1898. It has the atomic number of 84, was named after the country of Poland and given the symbol Po. Polonium-209 was found to emit alpha particles with a half-life of 102 years.
a) Write the nuclear equation that represents the radioactive decay of polonium-209 by alpha particle emission and identify the daughter isotope.
b) What is the decay constant of Po?
c) Supposedly we have a Po sample that has an activity of 6.660×109 disintegration per second stored in a 15.0 mL sealed container at 22°C for 235 days. How many alpha particles are formed inside the container? d) Polonium is the only element that adopts a simple cubic lattice. Given the density of Polonium is 9.32g/cm³, what is the radius of a Polonium atom?
Answer
a.)
Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle (α), which is a helium nucleus consisting of two protons and two neutrons. The result of alpha decay is the transformation of the original nucleus into a nucleus with an atomic number reduced by two and a mass number reduced by four.
In the case of polonium-209, the nucleus emits an alpha particle and transforms into a nucleus of lead-205, which has an atomic number of 82 and a mass number of 205. The nuclear equation for this decay is:
²⁰⁹Po → ⁴He + ²⁰⁵Pb
This equation shows that a polonium-209 nucleus decays by emitting an alpha particle, which is represented by the helium-4 nucleus. The daughter nucleus, lead-205, is shown on the right-hand side of the equation.
It is important to note that the sum of the atomic and mass numbers must be equal on both sides of the nuclear equation. This conservation of nuclear mass and charge is a fundamental principle of nuclear reactions.
Alpha decay is a common mode of decay for heavy and unstable nuclei such as polonium-209, which has a half-life of 102 years. This means that after 102 years, half of the initial number of polonium-209 atoms will have decayed into lead-205 atoms through alpha decay.
b.)
The decay constant (λ) is a fundamental property of a radioactive isotope that characterizes the probability of its decay. It is defined as the probability that an atom of the isotope will decay per unit time and is measured in units of inverse time, such as seconds⁻¹.
The half-life (t½) of an isotope is the time it takes for half of the original number of atoms in a sample to decay. It is an important parameter for understanding the rate of radioactive decay.
For Po-209, the half-life is given as 102 years. To calculate the decay constant, we first need to convert the half-life to seconds. There are 365 days in a year, 24 hours in a day, and 3600 seconds in an hour, so we can calculate the half-life in seconds as follows:
t½ = 102 years x 365 days/year x 24 hours/day x 3600 seconds/hour ≈ 3.22 × 10⁹ seconds
Now that we have the half-life in seconds, we can calculate the decay constant using the formula:
λ = ln(2) / t½
where ln(2) is the natural logarithm of 2, which is approximately 0.693. Substituting the values, we get:
λ = ln(2) / 3.22 × 10⁹ seconds ≈ 0.215 × 10⁻⁹ s⁻¹
Therefore, the decay constant of Po-209 is approximately 0.215 × 10⁻⁹ s⁻¹. This means that each Po-209 atom has a 0.215 × 10⁻⁹ s⁻¹ chance of decaying per second.
c.)
To calculate the number of alpha particles formed inside the container, we first need to understand the concept of radioactive decay and activity.
Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. The rate at which this decay occurs is known as the activity of the sample and is measured in disintegrations per second (dps) or Becquerels (Bq).
The activity of a radioactive sample decreases over time due to the decay of its radioactive nuclei. The rate of this decrease is proportional to the number of radioactive nuclei present in the sample and is given by the decay constant (λ).
The number of radioactive nuclei (N) in a sample at any given time can be expressed as:
N = N0e^(-λt)
where N0 is the initial number of radioactive nuclei at time t=0, t is the time elapsed since t=0, and e is the mathematical constant 2.718.
Now, let’s apply this concept to the given problem. We are given the activity (A) of the Po sample and the time (t) it has been stored in a sealed container. We need to calculate the number of alpha particles emitted during this time.
To do this, we use the formula:
N = (A / λ) × (1 – e^(-λt))
where λ is the decay constant of Po.
We are also given the volume (V) of the container and the density (ρ) of Po. These values can be used to calculate the number of Po atoms (n) in the sample as:
n = (V x ρ) / (atomic weight x Avogadro’s number)
where the atomic weight of Po is 209 g/mol and Avogadro’s number is 6.022 × 10^23/mol.
Since each Po atom emits one alpha particle during decay, the number of alpha particles formed inside the container is equal to the number of Po atoms in the sample.
Using the given values, we can calculate the answer as
To calculate the number of alpha particles formed inside the container, we can use the formula:
N = (A / λ) × (1 – e^(-λt))
where N is the number of Po atoms in the sample, A is the activity of the sample in disintegrations per second, λ is the decay constant, and t is the time the sample has been stored in seconds.
We are given: A = 6.660 × 10⁹
disintegrations per second V = 15.0 mL
t = 235 days x 24 hours/day x 60 minutes/hour x 60 seconds/minute ≈ 2.03 × 10⁷ seconds
The number of Po atoms in the sample can be calculated as:
N = (6.660 × 10⁹ / 0.215 × 10⁻⁹) × (1 – e^(-0.215 × 10⁻⁹ × 2.03 × 10⁷)) ≈ 5.64 × 10¹⁸ atoms
Since each Po atom emits one alpha particle during decay, the number of alpha particles formed inside the container is also equal to 5.64 × 10¹⁸.
d.)
To calculate the radius of a Polonium atom, we first need to understand the concept of a simple cubic lattice. A simple cubic lattice is a type of crystal lattice in which the lattice points are located at the corners of a cube. Each lattice point represents an atom or ion, and each cube contains one atom or ion. Polonium is the only element that adopts a simple cubic lattice, with each lattice point representing a polonium atom.
The formula for the radius of a simple cubic lattice is:
r = [(3M) / (4πρNₐ)]^(1/3)
where r is the radius of the atom, M is the atomic mass of the element, ρ is the density of the element, and Nₐ is Avogadro’s number.
In this case, we are given the atomic mass of Polonium-209 as M = 209 g/mol ≈ 209 × 10⁻³ kg/mol, the density of Polonium as ρ = 9.32 g/cm³ ≈ 9.32 × 10³ kg/m³, and Avogadro’s number as Nₐ = 6.022 × 10²³ atoms/mol.
Substituting the values into the formula, we get:
r = [(3 × 209 × 10⁻³) / (4π × 9.32 × 10³ × 6.022 × 10²³)]^(1/3)
Simplifying the equation, we get:
r = [(3 × 209 × 10⁻³) / (4 × 3.1416 × 9.32 × 10³ × 6.022 × 10²³)]^(1/3)
r = [(1.18 × 10⁻²) / (1.77 × 10²⁹)]^(1/3)
r ≈ 1.76 × 10⁻¹⁰ m
Therefore, the radius of a Polonium atom is approximately 1.76 × 10⁻¹⁰ m.
Summary
- The question asks us to write the nuclear equation for the radioactive decay of Polonium-209 by alpha particle emission and identify the daughter isotope. Polonium-209 emits an alpha particle, which is a helium nucleus, during decay. The daughter isotope produced is Lead-205.
- The question asks us to calculate the decay constant of Polonium-209 using its half-life. The half-life of Polonium-209 is 102 years, which we convert to seconds. Using the formula λ = ln(2) / t½, we calculate the decay constant of Polonium-209 to be approximately 0.215 × 10⁻⁹ s⁻¹.
- The question asks us to calculate the number of alpha particles formed inside a 15.0 mL sealed container storing a Polonium-209 sample with an activity of 6.660×10⁹ disintegration per second for 235 days at 22°C. Using the formula N = (A / λ) × (1 – e^(-λt)), we calculate the number of Polonium-209 atoms in the sample to be approximately 5.64 × 10¹⁸ atoms, and since each Po atom emits one alpha particle during decay, the number of alpha particles formed inside the container is also equal to 5.64 × 10¹⁸.
- The question asks us to calculate the radius of a Polonium atom given its density and atomic mass. Using the formula for the radius of a simple cubic lattice, r = [(3M) / (4πρNₐ)]^(1/3), we calculate the radius of a Polonium atom to be approximately 1.76 × 10⁻¹⁰ m.
FAQ:
Q: What is the decay constant of Polonium-209?
A: The decay constant of Polonium-209 is approximately 0.215 x 10^-9 s^-1.
Q: How can we calculate the decay constant of Polonium-209?
A: The decay constant can be calculated using the half-life of Polonium-209, which is 102 years, and the formula λ = ln(2) / t½, where t½ is the half-life.
Q: How can we calculate the number of alpha particles formed inside a container of Polonium-209?
A: The number of alpha particles formed inside a container of Polonium-209 can be calculated using the formula N = (A / λ) x (1 – e^-λt), where A is the activity of the sample, λ is the decay constant, and t is the time the sample has been stored.
Q: What is the radius of a Polonium atom?
A: The radius of a Polonium atom is approximately 1.76 x 10^-10 m. This can be calculated using the formula for the radius of a simple cubic lattice: r = [(3M) / (4πρNₐ)]^(1/3), where M is the atomic mass of Polonium-209, ρ is the density of Polonium, and Nₐ is Avogadro’s number.
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