Q1.
eqution for radioactive decay:
N=N0*exp(-lambda*t)
where N0=original number of atoms
labda=decay constant
N=number of undecayed atoms at time t
given that lambda=3.833*10^(-12)
original mass of carbon=0.21 kg/2=0.105 kg
let number of atoms of C14=x
then number of atoms of C12=10^12*x
total mass=((x/(6.023*10^23))*14)+((10^12*x/(6.023*10^23))*12)
grams
=1.9924*10^(-11)*x
==>1.9924*10^(-11)*x=105
==>x=5.27*10^12
so initial activity=x*decay constant
=20.2 Bq
part 2:
activity at present time=10.51 Bq
then N(t)=activity/lambda
=2.742*10^12
using it in decay equation:
2.742*10^12=5.27*10^12*exp(-lambda*t)
==>t=1.7046*10^11 seconds
=5405.2 years
Q3.
number of undecayed nucei in 3500 years in future=3500+5405.2
years from beginning
=8905.2 years
will be given by
N0*exp(-lambda*time)
=5.27*10^12*exp(-3.833*10^(-12)*8905.2*365*24*3600)
=1.7961*10^12
so activity=1.7961*10^12*3.833*10^(-12)
=6.8845 Bq
