What's the formula for radioactive decay?

The question is the half life of a radioactive chemical is 4 days. The initial mass of a sample is 2.5 grams, find (a) the mass that remains after 16 days, (b) the mass m(t) that remains after t days, (c) when will the mass be reduced to 0.05 grams.

2. Originally Posted by dm10
What's the formula for radioactive decay?

The question is the half life of a radioactive chemical is 4 days. The initial mass of a sample is 2.5 grams, find (a) the mass that remains after 16 days, (b) the mass m(t) that remains after t days, (c) when will the mass be reduced to 0.05 grams.
If you consider the amount of mass after 0, 4, 8, 12, 16 days etc. you'll see a pattern. That pattern is $m = (2.5) \left( \frac{1}{2} \right)^{t/4}$.

Use this rule (which is the answer to part b) to answer the other parts.

3. Originally Posted by dm10
What's the formula for radioactive decay?

The question is the half life of a radioactive chemical is 4 days. The initial mass of a sample is 2.5 grams, find (a) the mass that remains after 16 days, (b) the mass m(t) that remains after t days, (c) when will the mass be reduced to 0.05 grams.
$m_t = m_0 \left( {\frac{1}
{2}} \right)^{\frac{t}
{h}} \hfill \\$

${\text{where, }}m_0 = {\text{ initial mass }} = 2.5{\text{ g}} \hfill \\$

$m_t = {\text{ mass remained after time }}t \hfill \\$

$t = {\text{ time elapsed}} \hfill \\$

$h = {\text{ half - life period}} = 4{\text{ days}} \hfill \\$

$\left( a \right)\;\;t = 16{\text{ days}}{\text{.}} \hfill \\$

$m_t = 2.5\left( {\frac{1}
{2}} \right)^{\frac{{16}}
{4}} = 2.5\left( {\frac{1}
{2}} \right)^4 \hfill \\$

$= \frac{{2.5}}
{{16}} = 0.15625{\text{ g}} \hfill \\$

${\text{Mass remained after 16 days is 0}}{\text{.15625 grams}}{\text{.}} \hfill \\$

$\left( b \right)\;\;m_t = m_0 \left( {\frac{1}
{2}} \right)^{\frac{t}
{h}} \hfill \\$

$m_t = 2.5\left( {\frac{1}
{2}} \right)^{\frac{t}
{4}} \hfill \\$

${\text{Mass remained after t days is }}m_t = 2.5\left( {\frac{1}
{2}} \right)^{\frac{t}
{4}} {\text{ grams}} \hfill \\$

$\left( c \right)\;\;m_t = 0.05{\text{ g}}{\text{.}} \hfill \\$

$m_t = m_0 \left( {\frac{1}
{2}} \right)^{\frac{t}
{h}} \hfill \\$

$0.05 = 2.5\left( {\frac{1}
{2}} \right)^{\frac{t}
{4}} \hfill \\$

$\frac{{0.05}}
{{2.5}} = \left( {\frac{1}
{2}} \right)^{\frac{t}
{4}} \hfill \\$

$0.02 = \left( {0.5} \right)^{\frac{t}
{4}} \hfill \\$

${\text{Taking natural log on both sides,}} \hfill \\$

$\ln \left( {0.02} \right) = \ln \left[ {\left( {0.5} \right)^{\frac{t}
{4}} } \right] \hfill \\$

$\ln \left( {0.02} \right) = \frac{t}
{4}\ln \left( {0.5} \right) \hfill \\$

$t = \frac{{4 \times \ln \left( {0.02} \right)}}
{{\ln \left( {0.5} \right)}} = 22.57 \approx 23 \hfill \\$

$t = 23{\text{ days}} \hfill \\$

${\text{After 23 days the sample will remain 0}}{\text{.05 grams}}{\text{.}} \hfill \\$

4. Originally Posted by Shyam
$m_t = m_0 \left( {\frac{1}
{2}} \right)^{\frac{t}
{h}} \hfill \\$

${\text{where, }}m_0 = {\text{ initial mass }} = 2.5{\text{ g}} \hfill \\$

$m_t = {\text{ mass remained after time }}t \hfill \\$

$t = {\text{ time elapsed}} \hfill \\$

$h = {\text{ half - life period}} = 4{\text{ days}} \hfill \\$

$\left( a \right)\;\;t = 16{\text{ days}}{\text{.}} \hfill \\$

$m_t = 2.5\left( {\frac{1}
{2}} \right)^{\frac{{16}}
{4}} = 2.5\left( {\frac{1}
{2}} \right)^4 \hfill \\$

$= \frac{{2.5}}
{{16}} = 0.15625{\text{ g}} \hfill \\$

${\text{Mass remained after 16 days is 0}}{\text{.15625 grams}}{\text{.}} \hfill \\$

$\left( b \right)\;\;m_t = m_0 \left( {\frac{1}
{2}} \right)^{\frac{t}
{h}} \hfill \\$

$m_t = 2.5\left( {\frac{1}
{2}} \right)^{\frac{t}
{4}} \hfill \\$

${\text{Mass remained after 16 days is }}m_t = 2.5\left( {\frac{1}
{2}} \right)^{\frac{t}
{4}} {\text{ grams}} \hfill \\$

$\left( c \right)\;\;m_t = 0.05{\text{ g}}{\text{.}} \hfill \\$

$m_t = m_0 \left( {\frac{1}
{2}} \right)^{\frac{t}
{h}} \hfill \\$

$0.05 = 2.5\left( {\frac{1}
{2}} \right)^{\frac{t}
{4}} \hfill \\$

$\frac{{0.05}}
{{2.5}} = \left( {\frac{1}
{2}} \right)^{\frac{t}
{4}} \hfill \\$

$0.02 = \left( {0.5} \right)^{\frac{t}
{4}} \hfill \\$

${\text{Taking natural log on both sides,}} \hfill \\$

$\ln \left( {0.02} \right) = \ln \left[ {\left( {0.5} \right)^{\frac{t}
{4}} } \right] \hfill \\$

$\ln \left( {0.02} \right) = \frac{t}
{4}\ln \left( {0.5} \right) \hfill \\$

$t = \frac{{4 \times \ln \left( {0.02} \right)}}
{{\ln \left( {0.5} \right)}} = 22.57 \approx 23 \hfill \\$

$t = 23{\text{ days}} \hfill \\$

${\text{After 23 days the sample will remain 0}}{\text{.05 grams}}{\text{.}} \hfill \\$
Sometimes it can be educationally beneficial to leave something (apart from cutting and pasting) for the OP to do. That's why I replied in the way that I did.

If the OP needs more help s/he can always request it.

5. Thanks Shyam!

In contrast to Mr. Fantastic, your response definitely helped me a lot, and I, for one, definitely valued it as extremely helpful. I now know how to do radioactive decay!

6. Wow, this thing is still open? I posted this question over a year ago.

7. Originally Posted by dm10
Wow, this thing is still open? I posted this question over a year ago.
This is probably because this is a prime example of what not to do.

8. Meaning?

9. Originally Posted by dm10
Meaning?
Meaning not to come behind someone and give a complete and full solution. There is a certain amount of contempt held by this type offender for not only the original responder (in is case, Mr. F), but for the OP him/herself. This is Math Help Forum; not Math Answer Forum.

There is an ideal and moral standard that the true helpers try to maintain. And we live by this code, and we will die by this code.

OK, that was a little extreme, but it was illustriative of the notion in question.