If a 12.0-g sample of radon-222 decays so that after 19 days only 0.375 g remains, what is the half-life of radon-222?

1 Answer
Jun 26, 2017

The half life is =3.8 days

Explanation:

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The original mass is m_0=12.0g

After 19 days, amount remaining is m_19=0.375g

The ratio of m_19/m_0=0.375/12=1/32

That is

m_0/m_19=32

m_19=1/32m_0

The half life is =t_(1/2)

m_19 corresponds to 5t_(1/2)

So,

5t_(1/2)=19

t_(1/2)=19/5=3.8 days

We can perform the calculation

m_19=m_0e^(-19lambda)

e^(-19lambda)=m_19/m_0=1/32

19lambda=ln(32)

lambda=1/19ln(32)=5/19ln2

t_(1/2)=ln2/lambda=ln2/(5/19ln2)=19/5=3.8days