First slide

Radio activity

Question

1 mg gold undergoes decay with 2.7 days half- life period amount left after 8.1 days is

Easy

A

0.91 mg
B

0.25 mg
C

0.5 mg
D

0.125 mg

Solution

$N = N_{0} {(\frac{1}{2})}^{n} \Rightarrow N = N_{0} {(\frac{1}{2})}^{t / T_{1 / 2}} \Rightarrow N = 1 x {(\frac{1}{2})}^{\frac{8.1}{2.7}} = {(\frac{1}{2})}^{3} = \frac{1}{8} \Rightarrow N = \frac{1}{8} mg = 0.125 mg$

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