Question

The function f(h)=m(1/2)^h gives the mass, m, of a radioactive substance remaining after h half-lives. Iron has a half-life of 2.7 years. Which equation gives the mass of a 200 mg Iron sample remaining after y years, and approximately how many milligrams remain after 12 years?a) f(x) = 2.7(0.5)200; 1.7 mgb) f(x) = 200(0.5)12; 30.8 mgc )f(x) = 200(0.5)12; 30.8 mgd) f(x) = 200(0.185)12; 3.2 mg e)f(x) = 200(0.774)12; 9.2 mg

Answer 1

Answer: e)$f(x)=200(0.774)^{12}$; 9.2 mg

Step-by-step explanation:

Given function: $f(h)=m(\frac{1}{2})^h$

where m is the mass of a radioactive substance remaining after h half-lives.

 If iron has a half-life of 2.7 years, then for iron sample of mass =200 mg ,remain after y years the equation will be

$f(x)=200(\frac{1}{2})^{\frac{y}{2.7}}$

Now, to find mass remain after 12 years, substitute =12 in the equation , we get

$f(x)=200(\frac{1}{2})^{\frac{12}{2.7}}$

which is equivalent to

$f(x)=200(0.5)^{\frac{1}{2.7}\times12}\\\\\Rightarrow\ f(x)=200(0.774)^{12}$

On solving it we get,

$f(x)=200\times0.46\\\Rightarrow\ f(x)=9.2\ mg$

Therefore, the correct answer is e) $f(x)=200(0.774)^{12}$; 9.2 mg

Answer 2

The correct function should be: f(x) = 200(0.5)^(12/2.7)

This is an exponential equation, it is in the form of: y = ab^x
The a value is 200, of the starting mass.
The b value is the rate of change, which is 0.5.
The x value is the number of half-lives, which would be 12 divided by 2.7. 

If you evaluate the function, you would get about 9.2 mg.