Question

Question #6df28

Answer 1

Let `tau` be the half life.

By definition, after the first half life, the amount of radioactive substance should be half of the initial quantity.

Mathematically,

`(A(tau))/(A(0)) = 1/2`
or

`1/2 A(0) = A(tau)`.

Thus, proceed to solve the above equation for `tau`.

`1/2 xx 450 e^(-0.04 xx 0) = 450 e^(-0.04tau)`

`1/2 xx e^(0) = e^(-0.04tau)`

`1/2 = e^(-0.04tau)`

Take `ln` on both sides.

`ln(1/2) = ln(e^(-0.04tau))`

`-ln(2) = -0.04tau`

`tau = ln(2)/0.04`

`~~ 17.3`

Answer 2

Since you are trying to find the half-life of the substance and you know that `A` represents the final amount of the substance, set `A=450xx1/2=225`.

Setting `A` to equal `225` will allow you to find the half-life of the substance since you are assuming `225g` (half the amount of the substance) remains.

Thus:

`A(t)=450e^(-0.04t)`

`225=450e^(-0.04t)`

From this point on, we solve for `t`, the half-life of the substance.

Start by dividing both sides of the equation by `450`.

`color(red)(color(black)(225)/450)=color(red)((color(black)(450e^(-0.04t)))/450)`

`1/2=e^(-0.04t)`

Since the bases on both sides of the equation are not the same, take the natural logarithm of both sides.

`ln(1/2)=ln(e^(-0.04t))`

Using the natural logarithmic property, `ln_color(purple)b(color(red)m^color(blue)n)=color(blue)n*ln_color(purple)b(color(red)m)`, the equation simplifies into:

`ln(1/2)=-0.04t*ln(e)`

Solving for `t`, the half-life of the substance is:

`-0.04t=ln(1/2)/ln(e)`

`t=ln(1/2)/(-0.04ln(e))`

`t=ln(1/2)/(-0.04(1))`

`color(green)(|bar(ul(color(white)(a/a)t~~17.3color(white)(a/a)|)))`