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Answer to Question #25187 in Atomic and Nuclear Physics for Yonela

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
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.

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julius
26.02.13, 05:07

How 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