Question

6. Cesium-137 has a half-life of 30 years. Suppose a lab stores 30 mg in 1975. How much would be left in 2065? y = a (1 + r) (Fill in answer choices for a, r and t.)

Answer 1

The formula for calculating the amount remaining after a number of half years , n is :

`\begin{gathered} A=\frac{A_(\circ)}{2^n^{}} \\ \text{where A}_(\circ)\text{ =initial }amount \\ n=\frac{t}{t_{(1)/(2)}} \end{gathered}`

The lab store mass of Cesium-137 is 30mg in 1975

then the mass of Cesium-137 in 2065,

Time period =2065-1975

time period t=90 years,

substitute the value and solve for A

`\begin{gathered} A=\frac{30}{2^{(90)/(45)}} \\ A=(30)/(2^2) \\ A=(30)/(4) \\ A=7.5\text{ mg} \end{gathered}`

In 2065, the mass of Cesium -137 will be 7.5 mg

Answer : 7.5mg