Question

After 3 days a sample of radon-222 has decayed to 58% of its original amount.
(a) What is the half-life of radon-222? (Round your answer to two decimal places.)

Answer 1

please understand that whenever half-life is given or asked for,
the simplest equation to use is
f = 0.5^(t/h) , where f = fraction remaining , and h = half-life
=========
qa
t/h = ln f / ln 0.5
h = t(ln 0.5 / ln 0.58)
plugging in,
h = 3(ln 0.5 / ln 0.58) = 3.82 days <-------

qb
simlarly, t = h(ln f / ln 0.5)
plugging in,
t = 3.82(ln 0.1 / ln 0.5) = 12.68 days <-------

Answer 2

thanks

Answer 3

Your Welcome :)

Answer 4

theres actually another part to it
b) How long will it take the sample to decay to 10% of its original amount? (Round your answer to two decimal places.)

Answer 5

Well in general, for this question, the fomula for radioactive decay is:
\[m_f=m_i\left(\frac{1}{2}\right)^{\frac{t}{H}}\]
Where:
\[\eqalign{
&(m_f)\textit{ represents the final mass} \\
&(m_i)\textit{ represents the initial mass} \\
&(H)\textit{ represents the half life (time required for substance to halve in mass} \\
&(t)\textit{ represents the time taken since beginning}
}\]

So then we don't know our initial or final amount except that \(m_f=58\%\textit{ of }m_i=0.58m_i\)
We know our time is 3 days, which for practical reasons we will write is 36 hours
So then we can write:
\[0.58m_i=m_i\left(\frac{1}{2}\right)^\frac{36}{H}\]
We can divide by \(m_i\) to obtain:
\[\eqalign{
0.58&=\left(\frac{1}{2}\right)^\frac{36}{H} \\
\log_{10}(0.58)&=\frac{36}{H}\log_{10}\frac{1}{2} \\
\frac{\log(0.58)}{\log(0.5)}&=\frac{36}{H} \\
\frac{\log(0.58)}{36\log(0.5)}&=\frac{1}{H} \\
\frac{36\log(0.5)}{\log(0.58)}&=H
}\]

Answer 6

Whoops lol im a little slow there...ill do the second one!

Answer 7

By the way, I get 45.80 hours...

Answer 8

like for A I get 45.80 hours...any ideas @sana99 ?

Answer 9

And i'll do B right now @CrazyCarrie6655 :)

Answer 10

Assuming that the half life is 45.8 hours, then we have:
\[0.1=\left(\frac{1}{2}\right)^{\frac{t}{45.8}}\]
So we isolate for t and obtain:
\[45.8*\frac{\log0.1}{\log0.5}=t\approx152.14\]
So therefore, it will take 152.14 hours to decay to 10% it's original amount.

Answer 11

ok thanks

Answer 12

No prob :)