A particular radioactive isotope decays from 200 milligrams (mg) to 178 mg in 9 days. Find the half life of the isotope.?

1 Answer
Jan 26, 2018

The half life is #=53.5 " days"#

Explanation:

The initial mass is #m_0=200mg#

The mass after #9 " days"# is #m=178mg#

The equation for the radioactive decay is

#m=m_0e^(-lambdat)#

Where the radioactive constant is #=lambda#

Therefore,

#178=200e^(-9lambda)#

#178/200=e^(-9lambda)#

#e^(9lambda)=200/178#

#9lambda=ln(200/178)#

#lambda=1/9ln(200/178)#

But,

The half life is

#t_(1/2)=ln2/lambda#

So,

#t_(1/2)=ln2/(1/9ln(200/178))=53.5 " days"#