#### 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 *1*mg.

**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