Sodium-24 has a half-life of approximately 15 hours. If only one-eighth of the sodium-24 remains, about how much time has passed?

Question

a
30 hours
b
45 hours
c
60 hours
d
15 hours

Answer 1

To solve this problem, we can use the formula for exponential decay:

N(t) = N0 * (1/2)^(t/h)

Where:
N(t) is the amount of the substance remaining at time t
N0 is the initial amount of the substance
t is the time that has passed
h is the half-life of the substance

In this case, we know that only one-eighth of the sodium-24 remains, which means that N(t) = N0/8. We are trying to find out how much time has passed, so we can let t be the unknown variable. The half-life of sodium-24 is 15 hours, so we can substitute these values into the formula:

N(t) = N0 * (1/2)^(t/h)
N0/8 = N0 * (1/2)^(t/15)

Now we can simplify the equation:

(1/8) = (1/2)^(t/15)

To solve for t, we can take the logarithm of both sides of the equation:

log((1/8)) = log((1/2)^(t/15))

Using logarithm properties, we can bring down the exponent:

log((1/8)) = (t/15) * log((1/2))

Now we can solve for t:

8 = (2^(-t/15)) * log((1/2))

We can rewrite 8 as 2^3:

2^3 = 2^(-t/15) * log((1/2))

Since the bases are the same, we can set the exponents equal to each other:

3 = -t/15 * log((1/2))

We can rearrange this equation to solve for t:

t/15 = -3 / log((1/2))

Now we can solve for t:

t = -3 * 15 / log((1/2))

Using a calculator, we can evaluate this expression:

t ≈ -3 * 15 / (-0.6931)
t ≈ 45.46637502

Therefore, about 45 hours has passed. Answer choice b is the correct option.