If 1/8 of an original sample of krypton-74 remains unchanged after 34.5 minutes, what is the half life of krypton-74?

After the first half-life period, 1/2 of the sample remains.

After the 2nd period, only 1/4 remains.

After the 3rd period, only 1/8 remains.

Three half-life periods have been completed for 1/8 of the sample to remain, so 34.5/3 = 11.5 minutes.

M=m₀*2^ (-t/T)

m/m₀=2^ (-t/T)

ln (m/m₀) = - t*ln2/T

T=-t*ln2/ln (m/m₀)

T=-34.5min*ln2/ln0.125=11.5 min