I found an answer from math.stackexchange.com

Exponential **Decay** Question: The **Half**-**Life** of Cesium-137 is 30 ...

How do I find the relative growth **rate**? So after realizing that the **half** the amount
of 100mg would be 50mg I placed the 50 **mg** in my formula as the result. Thus,.

For more information, see Exponential **Decay** Question: The **Half**-**Life** of Cesium-137 is 30 ...

I found an answer from www.khanacademy.com

More exponential **decay** examples (video) | Khan Academy

A few more examples of exponential **decay**. Practice calculating k from **half**-**life**,
and calculating initial mass.

For more information, see More exponential **decay** examples (video) | Khan Academy

I found an answer from www.toppr.com

**Law** of **Radioactive Decay**, **Decay Rate**, **Half**-Mean **Life**, Q&A

In this topic, we will learn about the **Laws** of **Radioactive Decay**. ... Now,
integrating both the sides of the above **equation**, we get, ... This **rate** gives us the
number of **nuclei** decaying per unit time. ... Next, let's find the **relation between**
the mean **life** τ and the **disintegration constant** λ. ... Download **NCERT** Notes and
Solutions.

For more information, see **Law** of **Radioactive Decay**, **Decay Rate**, **Half**-Mean **Life**, Q&A

I found an answer from en.wikipedia.org

Nuclear fission product - Wikipedia

Nuclear fission products are the atomic fragments left after a large atomic nucleus
undergoes ... But ^{90}**Sr** has a 30-year **half**-**life**, and ^{89}**Sr** a 50.5-day **half**-**life**. ...
**rate** is highest for the shortest lived radionuclides, although they also **decay** the
fastest. ... Fission products have **half**-**lives** of **90 years** (samarium-151) or less,
except ...

For more information, see Nuclear fission product - Wikipedia

Required formulas:

Decay rate R = \lambda N where, N - number of radioactive atoms and \lambda - decay constant

Half life of the radioactive substance T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}

Given that

Half life of ^{90}_{38} Sr, T = 28 years

T =28*365*24*60*60 sec

T = 8.83 * 10^{8} sec

Mass of radioactive substance ( ^{90}_{38} Sr, m = 15 mg = 15 * 10^{-3} g )

Step 1: Finding the number of radioactive(Sr) atoms

Avogadro's number NA = 6.023 * 10^{23}

Atomic mass of strontium ^{90}_{38} Sr = 87.62 \approx 90

90g of strontium contains 6.023 * 10^{23} atoms.

So, Number of tomes in 15g of Sr, N = \frac{\text{NA}}{\text{Atomic mass}} * \text{ mass }

N = \frac{6.023* 10^{23}}{90} * 15 * 10^{-3}

N = 1.0038 * 10^{20} atoms

Step 2: Finding the decay rate of the radioactive substance(Sr)

R = \lambda N Half life T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}

\lambda = \frac{\ln 2}{T_{\frac{1}{2}}}

\lambda = \frac{0.693}{8.83 * 10^{8}}

Decay rate R = \lambda N

R = \frac{0.693}{8.83 * 10^{8}} * 1.0038 * 10^{20}

R = 7.9 * 10^{10} atoms/sec

Hence, decay rate of the 15 g of strontium R = 7.9 * 10^{10} atoms/sec