Answer:
See explanation
Explanation:
a) Formula for the mass of the sample that remains after t years = N= Noe^-kt
Where;
N = mass at time t years
No = mass at time t= 0
k = decay constant
t = time taken
So,
N = 100e^-kt
b) First,
t1/2 = -ln(1/2)/k
t1/2 = 0.693/k
t1/2 = half life of radium-226 =1590 years
1590 = 0.693/k
k = 0.693/1590
k = 4.36 * 10^-4
So,
N= 100e^-(4.36 * 10^-4 * 1000)
N= 65 mg
c) From
N = 100e^-kt
40 = 100e^-(4.36 * 10^-4t)
40/100 = e^-(4.36 * 10^-4t)
0.4 = e^-(4.36 * 10^-4t)
ln(0.4) = ln(e^-(4.36 * 10^-4t))
-0.9163 = -4.36 * 10^-4t
t = 0.9163/4.36 * 10^-4
t = 2102 years