Question

Could someone help me with this nuclear worksheet? The first part I feel sorta uneasy about, but I especially need help with 16-23. If you can, briefly explain how you got to your answer and/or show calculations.

Answer 1

16.

17. The relation between half life and decay constant is K= 0.693/ half life

K= 0.693/2.7=0.26 /day

From N= Noe^(-Kt)

where No is amount of sample at t=0 and N= sample at time t= 8.1 days

N = 100*e(^-0.26*8.1) =12.17 gms of sample is left

18.

from N =No e ^(-Kt)

t= 14.4 sec , N= 12.5 gms and No =50 gm

12.5 =50*e^(-14.4*K)

1/4= e^(-14.4*K)

-14.4K= ln(0.25)= -1.39

K= 1.39/14.4sec-1 =0.0965sec-1

Half life = 0.693/ K= 0.693/0.0965 sec=7.18 sec

19. N= 5gms , t= 40.35 day, Half life = 8.07 days, Decay constant = 0.693/half life = 0.693/8.07=0.086 days-1

N= No*e^(-Kt),   5= No* e(^{-0.086*40.35)}

No = 160.69 gm

20.

Decay constant = 0.693/ half life = 0.693/12.4=0.056 hr-1

From N= No*e^(-Kt)

N=750*e(-0.056*62)= 23.29 gms

21.

62.5= 500*e^(-K*639000)

0.125= e^(-K*639000)

ln (0.125)= -K*639000

K= 3.254*10^-6 yr-1

Half life = 0.693/ K= 0.693/ 3.254*10^-6 =212955 yrs

22.

Decay constant = 0.693/ half life = 0.693/5730 =0.000121 yr-1

0.137 A0 = Ao*e^{(-0.000121*t)}

0.137= e^{(-0.000121*t)}

ln (0.137)= -0.000121 t

the age of cloth = 16428 years

23. decay constant K= 0.000121yr-1

0.5801 A0 =Ao*e^{(-0.000121*t)}

t= 4500.45 years