Question

The half-life of cesium-137 is 30 years. Suppose we have a 20 mg sample.

a. Find the mass (in mg) that remains after t years.
b. How much of the sample (in mg) remains after 20 years?
c. After how long will only 1 mg remain?

Answer 1

Answer:

(a) A = (20mg)/(2^(t/30))

(b) 12.6mg

(c) 129.6years

Step-by-step explanation:

To calculate the amount remaining after a number of half-lives, n, we can make use of:

$A = \frac{B}{2^n}$

Where A = amount remaining

B = initial amount

$n = \frac{t}{t_{0.5}}$

(a) A = (20mg)/(2^(t/30))

(b) Mass after 20years

A = (20mg)/(2^(20/30)) ≈ 12.6mg

(c) After how long will only 1mg remain:

1mg = (20mg)/(2^(t/30))

$20mg = {2^{\frac{t}{30}}$

Taking log of both sides we have:

Log(20) = (t/30)log(2)

t/30 = (log(20))/(log(2)) ≈ 4.3

t/30 = 4.3

t = 30 x 4.3 ≈ 129.6years.