Diagnostic CP Q38

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NS_Tutor_Sophia
Posts: 30
Joined: Tue Aug 01, 2017 11:49 am

Diagnostic CP Q38

Post by NS_Tutor_Sophia » Wed Oct 17, 2018 9:23 am

A student recently asked us how to solve the following problem using the half-life equation: "Cobalt-60 has a half-life of 5.2 years. Starting with 32 grams of cobalt-60, how much of the original isotope will remain after 26 years?"

The half-life equation is N = N0(1/2)^(t/h) where N is the amount remaining, N0 is the original amount, t is the amount of time that has passed, and h is the half-life. Thus we obtain:

N = (32 g)(1/2)^(26 y/5.2 y)
N = (32 g)(1/2)^5
N = (32 g)(0.03125)
N = 1 g
Sophia Stone
PCAT Content Manager
Next Step Test Prep
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