To determine
(a)
To find: The mass of Cobalt-60.
Solution: ≈7.1.
Given: The half-life of Cobalt-60, t12=5.24years, initial mg m0=100, and m(5.24)=50
Formulae used: The standard exponential function
Explanation
Consider m=m0etk
m(5.24)=m0etkm(5.24)=100e5.24k
Hence, use the formulae of standard exponential function
m(5.24)=100e5.24k50100=e5.24ke5.24k=125.24k=ln(12)
Further simplify and get:
5.24k=ln(12)k=ln(12)5.24
Thus, the equation becomes m(5.24)=100e5.24(ln(12)5.24)
Now according to the given expression, the value of m(20)
m(5.24)=100e(ln(12)5.24)(5.24)m(20)=100e(ln(12)5.24)(20)≈7.1
Conclusion: Hence the value of mass is ≈7.1
(b)
To find: The time it takes for the mass to decay to 1mg.
Solution: ≈34.8years.
Given: The half-life of Cobalt-60, t12=5.24years, mt=1mg
Formulae used: The standard exponential function m(t)=100e5.24k
Consider m(t)=100e5.24k
Hence, use the formulae of standard exponential function
k=ln(12)5.24=−0.13228
Thus, the equation becomes m(t)=100e5.24(ln(12)5.24)t
Now according to the given expression, the value of m(1)=1
1=100e(−0.13228)t0.01=e−0.13228tln(0.01)=−0.13228tt=−ln(0.01)0.13228
Further simplify
t=−ln(0.01)0.13228≈34.8years
Conclusion: Hence the value of time it takes for the mass to decay is ≈34.8years