Question

1.Arsenic-74 is used to locate brain tumors. It has a half-life of 17.5 days. 90 mg wereused in a procedure. Write an equation that can be used to determine how much ofthe isotope is left after x number of half-lives.2. how much would be left after 70 days ?

Answer 1

`1)\text{ }N_t\text{ = 90\lparen}(1)/(2))^{(t)/(17.5)}`

2) 5.625 mg will be left

Step-by-step explanation:

1) Half-life = 17.5 days

initial amount of Arsenic-74 = 90 mg

To get the equation, we will use the equation of half-life:

`\begin{gathered} N_t\text{ = N}_0((1)/(2))^{\frac{t}{t_{(1)/(2)}}} \\ where\text{ N}_t\text{ = amount remaining} \\ N_0\text{ = initial amount} \\ t_{(1)/(2)\text{ }}\text{ = half-life} \end{gathered}`
`N_t\text{ = 90\lparen}(1)/(2))^{(t)/(17.5)}`

2) we need to find the remaining amount of Arsenic-74 after 70 days

t = 70

`\begin{gathered} N_t=\text{ 90\lparen}(1)/(2))^{(70)/(17.5)} \\ N_t\text{ = 5.625 mg} \end{gathered}`

So after 70 days, 5.625 mg will be left