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## Homework Statement

If 20% of a radioactive substance disappears in 70 days, what is its half-life?

## Homework Equations

y = C*e^(k*t)

where t is time in days

k is the constant of proportionality?

y is the current amount of substance

## The Attempt at a Solution

20% disappears, so 100% - 20% = 80% remaining = 0.80 after 70 days

.8*C = C*e^(70*k)

Solve for k

k = -3.188 * 10^-3

Plug k value back into equation to find t at the substance's half-life (.5)

.5*C = C*e^((-3.188 * 10^-3)*t)

Solve for t

t = 217.44

217.44 days is the substance's half-life.

Is that correct?

Thank you.