Measurements of a certain isotope tell you that the decay rate decreases from 8347decays/minute to?

Question

3049decays/minute over a period of 5.00days. What is the half-life T(1/2) of this isotope?

Answer 1

Final rate / Initial rate

=

(1/2) ^ (time / halflife)

Notice how when time = halflife, final rate is half the initial rate. That's why they call it the half life.

They give you final rate, initial rate, and time.

Solve for halflife.

halflife = time / (log (initial / final) / log (2))

Check my algebra. Then plug in the numbers.

Answer 2

Assume first-order decay.
C(t)= Co exp(-kt)
C(t)= 3049
Co = 8347
exp = 2.718
k = decay constant, day-1
t= time in days.(5)
C(t)/Co = exp(-kt), and solve for k (about 1/5).

To find half life, k t1/2 = 0.693. So t1/2 is about 3-1/2 days.