Question #0522c

1 Answer
Apr 23, 2017

Answer:

#Q(60.50" yrs") = 3.50xx10^-4" g"#

Explanation:

The general equation for exponential decay is:

#Q(t) = Q(0)e^(lambdat)#

Where Q(t) is the quantity at a given elapsed time, Q(0) is the initial quantity, and #lambda# is a decay coefficient

To find #lambda# given the half-life, you set #Q(t) = 1/2Q(0)#, t = the given time and then solve for #lambda#

#1/2Q(0) = Q(0)e^(lambda(5.27" yrs"))#

Divide both sides by Q(0):

#1/2 = e^(lambda(5.27" yrs"))#

To make the exponential function disappear, we use the natural logarithm:

#ln(1/2) = lambda(5.27" yrs")#

Replace #ln(1/2)# with -ln(2)

#-ln(2) = lambda(5.27" yrs")#

#lambda = -ln(2)/(5.27" yrs")#

Now we can evaluate at the equation at #t= 60.50" yrs"# and #Q(0) = 1" g"#

#Q(60.50" yrs") = (1" g")e^(-ln(2)/(5.27" yrs")60.50" yrs")#

#Q(60.50" yrs") = (1" g")e^(-ln(2)/(5.27" yrs")60.50" yrs")#

#Q(60.50" yrs") = 3.50xx10^-4" g"#