Question #25187

If the half-life for polonium-218 is 3.05 minutes. How many half-lives have passed for polonium-218 after 9.15 minutes
How much of an 89.6mg sample polonium-218 would remain after 9.15 minutes

Expert's answer

The half-life period for Po-218is t_hl = 3.05 minutes. Therefore, after t = 9.15 minutes, it has been (t / t_hl) = (9.15 min / 3.05 min) = 3 half-life periods for this substance. If the initial mass was 89.6 mg of Po-218, after 9.15 minutes the remainder

would be

m = m_0 * (1/2)^(t/t_hl) = 89.6 mg * (0.5)^(9.15 min / 3.05 min) = 89.6 mg *

(1/8) = 11.2 mg.

Answer: 3 half-life periods; 11.2 mg.

would be

m = m_0 * (1/2)^(t/t_hl) = 89.6 mg * (0.5)^(9.15 min / 3.05 min) = 89.6 mg *

(1/8) = 11.2 mg.

Answer: 3 half-life periods; 11.2 mg.

## Comments

julius26.02.13, 05:07How many half-lives have passed for polonium-218 after 9.15 minutes

Just divide 9.15 by 3.05 = 3. 3 half-lifes.

How much of an 89.6mg sample polonium-218 would remain after 9.15 minutes

Just divide the initial amount by 2 three times. 89.6/8 = 11.2mg

If you need to use the formula for decay (M2 = M1e^-ut) than solve for u first and then solve for M2

M2 = M1e^-ut

M1 = 1

M2 = 0.5

t=3.05x60

0.5 = e^-u(3.05x60)

ln(0.5) = -u(3.05x60)

u = 0.00378769

M2 = 89.6e^-0.00378769(9.15x60)

M2 = 11.2mg

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