Question

Part A

You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 26.3 minutes , what is the half-life of this substance?

Part B

An unknown radioactive substance has a half-life of 3.20 hours . If 10.6 g of the substance is currently present, what mass A0 was present 8.00 hours ago?

Answer 1

Part A

You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 26.3 minutes , what is the half-life of this substance?

Solution :-

Ln[A]t/[A]o = - k*t

Ln [100 /400] = - k * 26.3 min

-1.386 = - K * 26.3 min

-1.386 / -26.3 min = K

0.0527 min-1 = K

Now lets calculate the half life

T1/2 = 0.693 / K

        = 0.693 / 0.0527 min-1

       = 13.15 min

So the half life is 13.15 min

Part B

An unknown radioactive substance has a half-life of 3.20 hours . If 10.6 g of the substance is currently present, what mass A0 was present 8.00 hours ago?

Solution :- half life = 3.20 hr

Original = 10.6 g

Final = ?

Ln[A]t/[A]o = - k*t

Ln([A]t / 10.6 g) = - (0.693 / 3.20 hr) * 8.00 hr

Ln([A]t / 10.6 g) = -1.7325

[A]t / 10.6 g = anti ln [ -1.7325]

[A]t / 10.6 g = 0.1768

[A]t = 0.1768 * 10.6 g

[A]t = 1.87 g

So after 8 hr amount remain is 1.87 g