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Chapter 19: Problem 55

In the chapter, we saw that the unit curie corresponds to exactly \(3.70 \times 10^{10}\) nuclear disintegration per second for \(1 \mathrm{~g}\) of radium. Derive this unit given that the half-life of \({ }_{88}^{226} \mathrm{Ra}\) is \(1.6 \times 10^{3} \mathrm{yr}\).

Short Answer

Expert verified
The rate of decay for 1 g of radium-226, based on its half-life, is approximately \(3.66 \times 10^{10}\) disintegrations per second. This is very close to the given rate of \(3.70 \times 10^{10}\) disintegrations per second for a curie, confirming the calculation.

Step by step solution

01

Calculate the number of atoms in 1g of radium

Radium-226 has an atomic mass of approximately 226 amu. So, in 1 gram of radium, the number of atoms, 'N' is given by Avogadro's number divided by the atomic mass number. Hence, \(N = \frac{1 \text{ g}}{226 \text{ amu}} \times 6.022 \times 10^{23} \text{ atoms/gram mole} = 2.66 \times 10^{21} \text{ atoms}\)
02

Convert the half-life to seconds

The half-life of radium-226 is given in years. Convert this to seconds to match the units used in defining curie. There are approximately \(3.154 \times 10^{7}\) seconds in a year, so the half-life 'T' of radium-226 in seconds is \(T = 1.6 \times 10^{3} \text{ years} \times 3.154 \times 10^{7} \text{ s/year} = 5.05 \times 10^{10} \text{ s}\)
03

Apply the half-life decay formula

The rate of decay 'r' for a radioactive substance is given by the formula \(r = \frac{0.693 \times N}{T}\). Using the values for 'N' and 'T' calculated in the previous steps, \(r = \frac{0.693 \times 2.66 \times 10^{21} \text{ atoms}}{5.05 \times 10^{10} \text{ s}} = 3.66 \times 10^{10} \text{ disintegrations/s}\). This is very close to the given rate of \(3.70 \times 10^{10}\) disintegrations/s for a curie, confirming the calculation.

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Most popular questions from this chapter

Why do heavy elements such as uranium undergo fission while light elements such as hydrogen and lithium undergo fusion?

Tritium contains one proton and two neutrons. There is no proton-proton repulsion present in the nucleus. Why, then, is tritium radioactive?

As a result of being exposed to the radiation released during the Chernobyl nuclear accident, the dose of iodine-131 in a person's body is \(7.4 \mathrm{mCi}\) \(\left(1 \mathrm{mCi}=1 \times 10^{-3} \mathrm{Ci}\right) .\) Use the relationship rate \(=\lambda N\) to calculate the number of atoms of iodine- 131 to which this radioactivity corresponds. (The halflife of \({ }^{131} \mathrm{I}\) is 8.1 d. \()\).

Both barium (Ba) and radium (Ra) are members of Group \(2 \mathrm{~A}\) and are expected to exhibit similar chemical properties. However, \(\mathrm{Ra}\) is not found in barium ores. Instead, it is found in uranium ores. Explain.

Why is it preferable to use nuclear binding energy per nucleon for a comparison of the stabilities of different nuclei?
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