Problem 59 - College Physics

Question

An unstable isotope of cobalt, \(^{60} \mathrm{Co}\) , has one more neutron in its nucleus than the stable \(^{59}\mathrm{Co}\) and is a beta emitter with a half-life of 5.3 years. This isotope is widely used in medicine. A certain radiation source in a hospital contains 0.0400 \(\mathrm{g}\) of \(^{60} \mathrm{Co.}\) (a) What is the decay constant for that iso- tope? (b) How many atoms are in the source? (c) How many decays occur per second? (d) What is the activity of the source, in curies?

Step-by-Step Solution

Answer

(a) \(4.1504 \times 10^{-9} \text{s}^{-1}\); (b) \(4.014 \times 10^{20}\) atoms; (c) \(1.667 \times 10^{12}\) decays/s; (d) 45.05 Ci.

Step 1: Determine the Decay Constant

The decay constant, denoted by \( \lambda \), is related to the half-life \( t_{1/2} \) by the equation \( \lambda = \frac{\ln(2)}{t_{1/2}} \). Given \( t_{1/2} = 5.3 \) years, first convert this value into seconds by using the conversion \( 1 \text{ year} = 3.1536 \times 10^7 \text{ seconds}\):\[ t_{1/2} = 5.3 \times 3.1536 \times 10^7 \text{ seconds} = 1.670408 \times 10^8 \text{ seconds}. \]Then calculate the decay constant.\[ \lambda = \frac{\ln(2)}{1.670408 \times 10^8} \approx 4.1504 \times 10^{-9} \text{ s}^{-1}. \]

Step 2: Find the Number of Atoms

To find the number of atoms in the source, use Avogadro's number \( 6.022 \times 10^{23} \text{ atoms/mol} \) and the molar mass of \(^{60}\mathrm{Co}\) which is roughly 60 g/mol. Calculate the number of moles first.\[ \text{Moles of } ^{60}\mathrm{Co} = \frac{0.0400 \text{ g}}{60 \text{ g/mol}} = 6.67 \times 10^{-4} \text{ mol}. \]Then calculate the number of atoms.\[ N = 6.67 \times 10^{-4} \times 6.022 \times 10^{23} = 4.014 \times 10^{20} \text{ atoms}. \]

Step 3: Calculate Decays Per Second

Use the relationship \( \text{decays per second} = \lambda N \). From the previous steps, you have \( \lambda = 4.1504 \times 10^{-9} \text{ s}^{-1} \) and \( N = 4.014 \times 10^{20} \) atoms. Therefore,\[ \text{Decays per second} = \lambda N = 4.1504 \times 10^{-9} \times 4.014 \times 10^{20} = 1.667 \times 10^{12} \text{ decays per second}. \]

Step 4: Convert to Curies

1 Curie (Ci) is defined as \( 3.7 \times 10^{10} \) decays per second. Therefore, convert the activity in decays per second to Curies using:\[ \text{Activity (Ci)} = \frac{1.667 \times 10^{12}}{3.7 \times 10^{10}} = 45.05 \text{ Ci}. \]

Key Concepts

Decay Constant

In the world of radioactive decay, the decay constant, often symbolized by \( \lambda \), plays a crucial role. It represents the probability per unit time that a given nucleus will decay. The decay constant is directly linked to an isotope's half-life, which is the time required for half of a radioactive substance to decay. The formula to find the decay constant \( \lambda \) is straightforward:

• \( \lambda = \frac{\ln(2)}{t_{1/2}} \)

Here, \( \ln(2) \) is the natural logarithm of 2, approximately equal to 0.693, and \( t_{1/2} \) is the half-life of the isotope. This relationship shows that an isotope with a longer half-life will have a smaller decay constant, meaning it decays more slowly. Applying this concept to cobalt-60, with a half-life of 5.3 years, we convert the half-life to seconds and plug it into the formula to get the decay constant \( 4.1504 \times 10^{-9} \text{ s}^{-1} \). This tiny number illustrates the slow decay rate of this isotope.

Half-life Calculation

Half-life is a common measure used to describe how quickly unstable isotopes undergo radioactive decay. It is defined as the time taken for half of the radioactive nuclei in a sample to decay. For practical calculations, converting half-lives from years into seconds is often essential, especially when dealing with decay constants. To convert years to seconds, use:

• 1 year = \( 3.1536 \times 10^7 \) seconds

By multiplying the half-life in years by this conversion factor, you find the half-life in seconds. For the cobalt-60 isotope, this conversion results in a half-life of \( 1.670408 \times 10^8 \) seconds. Understanding half-life helps in determining how long a radioactive substance remains active and how frequently it emits radiation. This concept is especially vital in fields like medicine, where isotopes like cobalt-60 are used for radiation therapy.

Isotopes

Isotopes are different forms of the same element that have the same number of protons but different numbers of neutrons. This difference in neutron number leads to variations in atomic mass, but not in chemical properties. For example, cobalt-60 and cobalt-59 both have 27 protons, but cobalt-60 has 33 neutrons compared to the 32 neutrons in cobalt-59. Isotopes can be stable or unstable. Unstable isotopes undergo radioactive decay, a process where they emit radiation in an attempt to achieve stability. Cobalt-60, an unstable isotope, is a beta-emitter, releasing beta particles in its decay process. The presence of isotopes is widespread across many scientific fields, and their applications vary from nuclear medicine, where isotopes help in diagnostics and treatment, to industry, where they are used for material analysis and age dating.

Beta Emission

Beta emission is a type of radioactive decay where a beta particle is released from an atomic nucleus. It involves the transformation of a neutron into a proton, accompanied by the emission of an electron (beta-minus particle) and an antineutrino. This process increases the atomic number by one, transforming the element into a new isotope or a different element. In the case of cobalt-60, beta emission turns cobalt-60 into nickel-60, which is stable. The emitted beta particles can penetrate materials better than alpha particles, making them useful in medical applications like targeted radiotherapy, by damaging cancer cells while sparing surrounding healthy tissue. Beta radiation also finds a place in various industrial applications, such as quality control processes and material thickness measurements. Understanding beta emission is fundamental to grasp the transformation that takes place within an unstable isotope's nucleus and its impacts.