An igneous rock contains a Pb−206/U−238 mass ratio of 0.367.
How old is the rock? (U−238 decays into Pb−206 with a half-life of 4.50×109 yr.)
I'm not sure how to get started on this :/
I don't know how advanced this class is but there is an easy way (and not quite so accurate) and a harder way (but a little more accurate).
Easy way.
You start with 100 atoms U and they decay over the years to form x atoms Pb which leaves 100-x atoms U. The ratio of Pb/U is 0.367 so substitute
(x/100-x) = 0.367 and solve for x and call that grams U. You should get approx 27g U but you can work it out more accurately if you wish. This is a close approximation. Then use
ln(No/N) = kt
No = 100
N = 100-27 = approx 73
k = 0.693/t1/2 from the problem. Solve for t. If I didn't make an error the answer is 2.02E9 years for the age of the rock. This is the easy way. What's wrong with it? We have used atoms U decaying to atoms of Pb (which is ok) BUT we have used a MASS RATIO and not an atom ratio; i.e., U weighs 238 g/mol and Pb only weighs 206 g/mol so can't really equation atoms with grams.
The longer but a little more accurate way. Convert atoms U to grams and atoms Pb to grams, then we can use the mass ratio without any problems. It's a little messier math problem but here is how you do it.
U initially = 100 atoms. mass 100 atoms is (238*100/6.022E23) = ?
x atoms decay. They have a mass of (238x/6.022E23)
mass U remaining is the difference. I'll let you do that.
They decay to form Pb. Those atoms of x will have a mass of (206x/6.022E23)
Substitute mass Pb/mass U and solve for x. I obtained approx 30g but I'll let you do the math.
Then ln(No/N) = kt
No = 100
N = approx 70
k = same as the first above
t = solve and I obtained about 2.29E9 years. If you count years difference between 2.02E9 and 2.29E9 that's a bunch of years(something like 2.7E8 or 270,000,000 years) but it's small compared to the time scale. Only about 10% error. So doing it the easy way saves a lot of time and doesn't make that much error.
Well, it sounds like you've stumbled upon a rock-solid question there! Let's break it down and have some fun with it.
The Pb-206/U-238 mass ratio of 0.367 tells us that for every 0.367 grams of lead-206 (Pb-206), we have 1 gram of uranium-238 (U-238).
Now, since uranium-238 is decaying into lead-206, we can use its half-life of 4.50×10^9 years to determine how much time has passed.
So, here's the joke: Why did the rock go on a date with its radioactive friend? Because it needed some time to decay! Ba-dum-tss!
Alright, back to business. We can set up an equation using the half-life formula:
mass of U-238 / mass of Pb-206 = e^(−kt)
In this equation, k is the decay constant, and t is the time in years. We're trying to find t.
Now, since the mass ratio is 0.367, we can set up the equation:
0.367 / 1 = e^(−k * 4.50×10^9)
To solve for k, we need some math magic. Take the natural logarithm of both sides of the equation:
ln(0.367) = −k * 4.50×10^9
Now, divide both sides by -4.50×10^9:
k = ln(0.367) / -4.50×10^9
With the value of k in hand, we can substitute it back into the equation:
0.367 / 1 = e^(k * t)
But we're trying to solve for t, so we need to rearrange the equation:
t = ln(0.367) / (k * -1)
Now, plug in the value of k and crunch the numbers to find the age of the rock!
And voila! You'll have the age of the rock faster than you can say, "Pebbles, stones, and boulders, oh my!"
Remember, science can be a rocky road, but with a little humor, we can make it smooth sailing!
To find the age of the rock, we can use the concept of radioactive decay and the information given.
The decay of U-238 into Pb-206 can be described by the equation:
U-238 -> Pb-206 + 8 He-4
The decay constant λ can be calculated using the half-life of U-238:
λ = ln(2) / t₁/₂
= ln(2) / (4.50×10⁹ yr)
≈ 0.154 yr⁻¹
The decay constant λ gives the rate of decay for U-238. We can use the equation for radioactive decay to relate the initial abundance of U-238 to the current abundance of Pb-206:
(Pb-206/U-238) = e^(λ t)
Where (Pb-206/U-238) is the mass ratio of Pb-206 to U-238, e is the base of the natural logarithm, λ is the decay constant, and t is the age of the rock.
Now, let's solve for t:
0.367 = e^(0.154 t)
We can solve this equation by taking the natural logarithm on both sides:
ln(0.367) = ln(e^(0.154 t))
Using the properties of logarithms, we can simplify the equation:
ln(0.367) = 0.154 t
Now, solve for t:
t = ln(0.367) / 0.154
Using a calculator, we can evaluate this expression:
t ≈ -2.345 / 0.154
t ≈ -15.20
Since time cannot be negative, we discard the negative solution. Therefore, the age of the rock is approximately 15.20 billion years.
To determine the age of the rock, we can use the concept of radioactive decay. In this case, we are dealing with the decay of Uranium-238 (U-238) into Lead-206 (Pb-206). The half-life of U-238 is given as 4.50×10^9 years.
To calculate the age of the rock, we need to use the decay equation:
N(t) = N₀ * (1/2)^(t/T)
Where:
- N(t) is the current amount of radioactive substance (Pb-206) after time t
- N₀ is the initial amount of the radioactive substance (Pb-206) at the beginning
- t is the elapsed time
- T is the half-life of the radioactive substance (U-238)
In our case, we are given the mass ratio of Pb-206/U-238 in the rock, which signifies the ratio of Pb-206 to U-238 atoms. To simplify, let's assume we have 1 gram of the rock. Therefore, the mass of U-238 would be 0.367 grams.
Now, we can use the atomic masses of Pb-206 and U-238 to convert the mass of U-238 into the number of atoms. The atomic mass of U-238 is 238 grams/mol, and Pb-206 is 206 grams/mol. So, there are (0.367/238) * Avogadro's number of U-238 atoms in the rock.
Once we have the number of U-238 atoms, we can find the number of Pb-206 atoms using the decay equation. Since the ratio of Pb-206 to U-238 atoms is constant throughout the decay process, we can say:
N(Pb-206) = N₀(Pb-206) * (1/2)^(t/T)
Since we already assumed 1 gram of the rock, N₀(Pb-206) is equal to 0.367 * Avogadro's number.
Finally, to find the age of the rock (t), we need to solve the equation for t. Rearranging the equation:
t = - T * (log₂(N(Pb-206) / N₀(Pb-206)))
Substituting the known values, such as N(Pb-206), N₀(Pb-206), and T, we can calculate the age of the rock.
It's worth noting that this calculation assumes that the rock initially contained no lead-206 when it formed, and all the lead-206 present in the rock came from the decay of uranium-238. This is a simplified assumption, and other factors like contamination or other sources of lead can influence the age determination.