A radioactive substance decays exponentially in such a way that after 50 years, 60% of the initial amount remains. Find an expression for the quantity remaining after t years.
To find an expression for the quantity remaining after t years, we can use the formula for exponential decay.
The formula for exponential decay is given by:
N = N₀ * e^(-kt),
where:
- N is the quantity remaining after t years,
- N₀ is the initial quantity,
- e is the base of the natural logarithm (approximately 2.71828),
- k is the decay constant, which determines the rate of decay,
- t is the time in years.
In this problem, we are given that after 50 years, 60% of the initial amount remains. Let's use this information to find the decay constant, k.
We know that:
N = N₀ * e^(-50k) = 0.6N₀.
Dividing both sides by N₀ gives us:
e^(-50k) = 0.6.
To find k, we need to take the natural logarithm (ln) of both sides:
ln(e^(-50k)) = ln(0.6).
Using the property of logarithms that ln(e^x) = x, we get:
-50k = ln(0.6).
Now, we can solve for k by dividing both sides by -50:
k = -ln(0.6)/50.
Now that we found the value of k, we can substitute it back into the equation N = N₀ * e^(-kt) to get the expression for the quantity remaining after t years:
N = N₀ * e^(-(-ln(0.6)/50)t).
Simplifying this expression further, we get:
N = N₀ * e^(ln(0.6)/50t).
Finally, we can simplify it to:
N = N₀ * (0.6)^(1/50t).
Therefore, the expression for the quantity remaining after t years is N = N₀ * (0.6)^(1/50t).
R(t) = e^kt, where R(t) is the percentage remaining after t t years, k is a constant
given: when t = 50 , R(50) = .6
.6 = e^(50k)
ln both sides
ln.6 = 50k lne
k = ln.6/50
R(t) = e^( (ln.6/50) t)