Half life of bromine −82=36 hours
t1/2=K0.693
K=360.693=0.01925hr−1
K=t2.303logNtN0
(t=1day=24hr)
logNtN0=2.30324hr×0.01925hr−1
=0.2006
OrNtN0= anti log (0.2006)
NtN0=1.587
Or N0Nt=1.5871=0.630
Hence, the fraction remain after day =63×10−2