Iodine-131 is a radioactive isotope with a half-life of 8 days. How many grams of a 64 g sample of iodine-131 will remain at the end of 24 days?

2 Answers
May 25, 2016

Answer:

#"8 g"#

Explanation:

The nuclear half-life of a radioactive isotope tells you how much time must pass in order for half of the atoms present in an initial sample to undergo radioactive decay.

In essence, the half-life tells you at what time intervals you can expect an initial sample of a radioactive isotope to be halved.

http://ch302.cm.utexas.edu/nuclear/radioactivity/selector.php?name=half-life

In your case, iodine-131 is said to have a half-life of #8# days. This means that with every #8# days that pass, your sample of iodine-131 will be halved.

If you take #A_0# to be the initial sample of iodine-131, you will be left with

#1/2 * A_0 = A_0/2 -> # after the passing of one half-life

#1/2 * A_0/2 = A_0/4 -># after the passing of two half-lives

#1/2 * A_0/4 = A_0/8 -># after the passing of three half-lives
#vdots#

and so on.

You can thus say that the amount of iodine-131 that remains undecayed, #"A"#, after a period of time #t#, will be equal to

#color(blue)(|bar(ul(color(white)(a/a)A = A_0 * 1/2^ncolor(white)(a/a)|)))#

Here #n# represents the number of half-lives that pass in #t#.

In your case, you have

#n = (24 color(red)(cancel(color(black)("days"))))/(8color(red)(cancel(color(black)("days")))) = 3#

This will get you

#A_"24 days" = "64 g" * 1/2^3#

#A_"24 days" = color(green)(|bar(ul(color(white)(a/a)"8 g"color(white)(a/a)|)))#

May 25, 2016

Answer:

#=> m_t = 8g#

Explanation:

Half Life follows first Order kinetics

enter image source here

This graph shows what is going on during the decay

As you can see from the graph the half life is 8 days

Lets do the math and verify it with the graph

Recall

# t_(1/2) = ln2/k#

#8 = ln2/k#

#k = ln2/8#

With this you can solve most problems regarding the above data

#ln(A_o/A_t) = kt#

#ln(A_o/A_t) = ln2/8 * 24 = 3 ln2 = ln 2^3 = ln8#

#(A_o/A_t) = 8#

Now

Greater the concentration of a sample greater the mass in in it

#C prop M#

So

#(A_o/A_t) = (m_o/m_t) = 8#

#"we have "m_o = 64 g"#

#64/m_t = 8#

#=> m_t = 8g#

Verifying with the graph

#m_("24 hours") = 12.5 % m_o = 12.5 *10^-2 *m_o#
# =12.5 *10^-2 *64 = 8g#

Hence verified