Question #6df28

2 Answers
Apr 5, 2016

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#

Apr 5, 2016

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)|)))#