The half-life for the radioactive decay of C-14 is 5730 years.How long will it take for 30% of the C-14 atoms in a sample of C-14 to decay?

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Answer 1 · 2017-12-17T08:13:46

The time is #=2948.5y#

The half life of #C-14# is

#t_(1/2)=5730y#

The radioactive constant is

#lambda=ln2/t_(1/2)=ln2/5730=1.21*10^-4y^-1#

Apply the equation

#m(t)=m_0e^(lambdat)#

#(m(t))/m_0=e^(lambdat)#

#lambdat=ln((m(t))/m_0)#

#t=1/lambdaln((m(t))/m_0)#

The time is

#t=1/(1.21*10^-4)*ln(0.7)=2948.5y#