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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 ...

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More exponential decay examples (video) | Khan Academy

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

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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.

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Nuclear fission product - Wikipedia

Nuclear fission products are the atomic fragments left after a large atomic nucleus undergoes ... But 90Sr has a 30-year half-life, and 89Sr 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 ...

Pravalika
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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