The rubidium isotope 87Rb is a β emitter that has a half-life of 4.9 x 1010 y that decays into 87Sr. It is used to determine the age of rocks and fossils. Rocks containing the fossils of early animal have a ratio of 87Sr to 87Rb of 0.004. Assuming that there was no 87Sr present when the rocks were formed, calculate the age of the fossils.
N=N₀•exp(-ln2•t/T)
N/N₀=1-0.004=0.996
ln0.996= - ln2•t/4.9•10¹º
t= ln0.996•4.9•10¹º/ln2= 2.83•10⁸ years
Oh, the age of fossils, huh? That's a tricky one! Let's see if I can wrap my clown brain around it.
We know that rubidium-87 (87Rb) decays into strontium-87 (87Sr) with a half-life of 4.9 x 10^10 years. And we have a rock with a 87Sr to 87Rb ratio of 0.004. Interesting!
Now, assuming there was no 87Sr present when the rocks were formed, we can calculate the age of the fossils by looking at how much 87Rb has decayed into 87Sr. Are you ready? Here we go!
To find the age, we can use the decay equation:
N(t) = N₀ * (1/2)^(t/t₁/₂)
Where:
N(t) is the current amount of 87Rb (which we're assuming to be 0 since it has all decayed),
N₀ is the initial amount of 87Rb (which we're assuming to be whatever the ratio is multiplied by the total amount of rubidium),
t is the age we're trying to find,
and t₁/₂ is the half-life of 87Rb.
Now, since we know the ratio is 0.004, and the 87Rb to 87Sr ratio decreases by half every half-life, we can set up this equation:
0.004 = 1 * (1/2)^(t/4.9 x 10^10)
Now, let's solve that bad boy and see how old those fossils are!
Calculating... calculating... carry the one...
Hold on to your funny bones, we have an answer!
The age of the fossils is approximately 1.6 x 10^11 years. Ta-dah!
Please note that this is just a rough estimation, and I can't guarantee accuracy. But hey, it's all in good clownish fun!
To calculate the age of the fossils, we can use the radioactive decay equation:
N(t) = N₀ * (1/2)^(t / τ)
Where:
N(t) = the number of radioactive atoms at time t
N₀ = the initial number of radioactive atoms
t = time elapsed
τ = half-life of the radioactive element
In this case, the radioactive element is 87Rb, and we know the half-life (τ) is 4.9 x 10^10 years.
We are given the ratio of 87Sr to 87Rb in the rocks containing the fossils, which is 0.004. This ratio corresponds to the number of 87Sr atoms (N₁) to 87Rb atoms (N₀) when the rocks formed.
N₁ / N₀ = 0.004
Since there was no 87Sr present initially, N₀ = N(t).
Therefore, the equation becomes:
N₁ / N(t) = 0.004
Now we rearrange the equation to solve for t:
N(t) = N₁ / 0.004
Substituting the decay equation, we have:
N₀ * (1/2)^(t / τ) = N₁ / 0.004
Since N₀ = N(t), we can simplify further:
N(t) * (1/2)^(t / τ) = N₁ / 0.004
Now we substitute the values given:
N(t) * (1/2)^(t / (4.9 x 10^10)) = N₁ / 0.004
If we divide both sides of the equation by N(t), we get:
(1/2)^(t / (4.9 x 10^10)) = N₁ / (0.004 * N(t))
The left side of the equation represents the fraction of the original radioactive atoms remaining after time t, and the right side represents the fraction of 87Rb atoms that have decayed to form 87Sr.
We are given the ratio N₁ / N(t) = 0.004, so the equation becomes:
(1/2)^(t / (4.9 x 10^10)) = 0.004
To solve for t, we can take the logarithm (base 2) of both sides:
t / (4.9 x 10^10) = log₂(0.004)
Now, multiply both sides by (4.9 x 10^10):
t = (4.9 x 10^10) * log₂(0.004)
Calculating this using a calculator, we find:
t ≈ 1.7 x 10^10 years
Therefore, the age of the fossils is approximately 1.7 x 10^10 years.
To calculate the age of the fossils, we can use the concept of radioactive decay and the decay equation.
The decay equation for a radioactive isotope is given by:
N(t) = N(0) * e^(-λt)
where N(t) is the number of radioactive atoms remaining at time t, N(0) is the initial number of radioactive atoms, λ is the decay constant, and e is the base of the natural logarithm.
In this case, we can use the ratio of 87Sr to 87Rb to determine the initial number of 87Rb atoms when the rocks were formed.
Given that the ratio of 87Sr to 87Rb is 0.004, we can calculate the initial number of 87Rb atoms by dividing the ratio by the natural abundance of 87Rb (0.278%).
Initial number of 87Rb atoms (N(0)) = 0.004 / 0.00278 ~= 1.439
Now, we need to find the decay constant (λ) for the 87Rb isotope.
The decay constant (λ) is related to the half-life (t1/2) by the equation:
λ = ln(2) / t1/2
Given that the half-life of 87Rb is 4.9 x 10^10 years, we can calculate the decay constant:
λ = ln(2) / (4.9 x 10^10) = 1.416 x 10^-11 yr^-1
Now, we can substitute the values into the decay equation and solve for the age (t) of the fossils:
N(t) = N(0) * e^(-λt)
1 = 1.439 * e^(-1.416 x 10^-11 * t)
Divide both sides by 1.439:
1 / 1.439 = e^(-1.416 x 10^-11 * t)
Take the natural logarithm of both sides:
ln(1 / 1.439) = ln(e^(-1.416 x 10^-11 * t))
Use the inverse natural logarithm (exponential function) to solve for t:
t = ln(1 / 1.439) / (-1.416 x 10^-11)
Using a calculator or software, evaluate the expression:
t = ln(0.6942) / (-1.416 x 10^-11)
t ≈ 4.38 x 10^10 years
Therefore, the age of the fossils is approximately 4.38 x 10^10 years.