Question

Technetium-99m is used as a radioactive tracer for certain medical tests. It has a half-life of 1 day. Consider the function T where T (d) = 100(2)^-2 is the percent of Technetium-99m remaining d days after the tests. Write an expression that represents the number of days until only 10% remains.

Answer 1

Answer:

The expression that represents the number of days until only 10% remains is T((d) 10 %) =100×$(\frac{1}{2} )^{3.322}$.

Step-by-step explanation:

The equation for half life is of the form

A = A₀×$(\frac{1}{2} )^{\frac{t}{h} }$.........................................................................(1)

Where

A = Final amount

A₀ = Initial amount

t = Time

h = Half life

For the equation T(d) = 100×2⁽⁻²⁾....................................(2)

We have by comparison with the equation for half life

2 ≡ $\frac{t}{h}$  and and the equation (2) can be written as

Percentage remaining after 2 half lives is

$\frac{A}{A_0}$ ×100=100×$(\frac{1}{2} )^{2 }$

However if the half life of Technetium-99m is 6 hours then we have for one day

$\frac{A}{A_0}$ ×100=100×$(\frac{1}{2} )^{2 *2}$

Therefore an expression that represents the number of days until only 10% remains is

$\frac{A}{A_0}$ ×100=100×$(\frac{1}{2} )^{\frac{d}{h} }$ = 10 %

$(\frac{1}{2} )^{\frac{d}{h} }$ =$\frac{1}{10}$

= ㏑$(\frac{1}{2} )^{\frac{d}{h} }$  = ㏑($\frac{1}{10}$)

= $\frac{d}{h}$×㏑$(\frac{1}{2} )$ = ㏑($\frac{1}{10}$)

$\frac{d}{h}$ = $\frac{ln(\frac{1}{10}) }{ln(\frac{1}{2} )}$ = 3.322

Therefore the expression for the number of days 10 % of Technetium-99m will be remaining is

T((d) 10 %) =100×$(\frac{1}{2} )^{3.322}$