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

$1)\text{ }N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}$

2) 5.625 mg will be left

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

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

t = 70

$\begin{gathered} N_t=\text{ 90\lparen}\frac{1}{2})^{\frac{70}{17.5}} \\ N_t\text{ = 5.625 mg} \end{gathered}$

So after 70 days, 5.625 mg will be left