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The amount of a radioactive substance remaining after [tex]\( t \)[/tex] years is given by the function [tex]\( f(t) = m(0.5)^{\frac{t}{n}} \)[/tex], where [tex]\( m \)[/tex] is the initial mass and [tex]\( n \)[/tex] is the half-life in years. Cobalt-60 has a half-life of about 5.3 years.

Which equation gives the mass of a 50 mg Cobalt-60 sample remaining after 10 years, and approximately how many milligrams remain?

A. [tex]\( f(10) = 50(0.5)^{\frac{10}{5.3}} ; 13.5 \)[/tex] mg
B. [tex]\( f(10) = 50(0.50)^{0.53} ; 34.6 \)[/tex] mg
C. [tex]\( f(10) = 5.3(0.5)^5 ; 0.2 \)[/tex] mg
D. [tex]\( f(10) = 5.3(0.5)^{0.2} ; 4.6 \)[/tex] mg



Answer :

GinnyAnswer
To determine the amount of Cobalt-60 remaining after 10 years, we need to use the given function [tex]\( f(t) = m(0.5)^{\frac{t}{n}} \)[/tex], where [tex]\( m \)[/tex] is the initial mass and [tex]\( n \)[/tex] is the half-life in years.

Given:
- Initial mass ([tex]\( m \)[/tex]) = 50 mg
- Half-life ([tex]\( n \)[/tex]) of Cobalt-60 = 5.3 years
- Time ([tex]\( t \)[/tex]) = 10 years

We need to substitute these values into the function:

[tex]\[ f(t) = 50 \left( 0.5 \right)^{\frac{10}{5.3}} \][/tex]

Firstly, compute the exponent:

[tex]\[ \frac{10}{5.3} \approx 1.8868 \][/tex]

Thus, the equation is:

[tex]\[ f(10) = 50 \left( 0.5 \right)^{1.8868} \][/tex]

The equation we formed is [tex]\( f(10) = 50(0.5)^{\frac{10}{5.3}} \)[/tex]. Now, calculating the remaining amount:

[tex]\[ f(10) \approx 50 \left( 0.5 \right)^{1.8868} \approx 13.52 \][/tex]

Therefore, the correct equation and remaining mass of the Cobalt-60 sample after 10 years are:

[tex]\[ f(10) = 50(0.5)^{\frac{10}{5.3}}; \approx 13.5 \text{ mg} \][/tex]

So, the correct answer in the options provided is:

[tex]\[ f(10)=50(0.5)^{\frac{10}{5.3}} ; 13.5 \text{ mg} \][/tex]

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