Problem 55 - College Physics

Question

A sample of \({ }^{215} \mathrm{Bi}\), which beta decays \(\left(t_{1 / 2}=2.4 \mathrm{~min}\right)\) initially contains one-hundreth of Avogadro's number of nuclei. (a) What is the sample's mass? (b) Write down the beta decay equation and predict the product nucleus. (c) How many bismuth nuclei are present after \(10 \mathrm{~min} ?\) (d) After \(1.0 \mathrm{~h} ?\) (d) What are the activities, in curies and becquerels, at these times?

Step-by-Step Solution

Answer

(a) 2.15 g; (b) \(^{215}_{84} \mathrm{Po}\) + \(\beta^-\); (c) \(4.17 \times 10^{19}\); (d) 0 nuclei; activ.: 2.02x10^{17} Bq & 5.45 Ci at 10 min; 0 at 1 hr.

Step 1: Determine Initial Number of Nuclei

Avogadro's number is approximately \(6.022 \times 10^{23}\). Since our sample contains one-hundredth of Avogadro's number of \(^{215}\mathrm{Bi}\) nuclei, we have \(\frac{6.022 \times 10^{23}}{100} = 6.022 \times 10^{21}\) nuclei to start with.

Step 2: Calculate Sample's Mass

The mass of one mole of \(^{215}\mathrm{Bi}\) is 215 grams. From Step 1, we know we have \(6.022 \times 10^{21}\) nuclei. Therefore, the mass \(m\) of the sample is given by \[ m = \left(\frac{6.022 \times 10^{21}}{6.022 \times 10^{23}}\right) \times 215 \text{ grams} = 2.15 \text{ grams}. \]

Step 3: Write Beta Decay Equation

Beta decay transforms \(^{215}\mathrm{Bi}\) into another nucleus. During beta decay, a neutron is converted into a proton, electron, and antineutrino. The decay equation is:\[ ^{215}_{83} \mathrm{Bi} \rightarrow \, ^{215}_{84} \mathrm{Po} \,+ \, \beta^- + \, \bar{v}_{e} \] The product nucleus is polonium-215.

Step 4: Calculate Number of Nuclei After 10 minutes

The decay of nuclei over time can be calculated using the formula:\[ N(t) = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}} \]Where \(t\) is the time, \(t_{1/2}\) is the half-life (2.4 minutes), and \(N_0\) is the initial number of nuclei. After 10 minutes,\[ N(10) = 6.022 \times 10^{21} \left(\frac{1}{2}\right)^{10/2.4} \approx 4.17 \times 10^{19} \] nuclei remain.

Step 5: Calculate Number of Nuclei After 1 Hour

Convert 1 hour to minutes (60 minutes) and apply the decay formula:\[ N(60) = 6.022 \times 10^{21} \left(\frac{1}{2}\right)^{60/2.4} \approx 0 \] nuclei, effectively no \(^{215}\mathrm{Bi}\) nuclei remain after 1 hour.

Step 6: Determine Activities at 10 minutes and 1 hour

Activity is the rate of decay, given by \(A = \lambda N\), where \(\lambda = \frac{\ln(2)}{t_{1/2}}\). First, calculate \(\lambda\):\[ \lambda = \frac{\ln(2)}{2.4 \, \text{min}} = 0.289 /\text{min} \]**At 10 minutes:** \[ A(10) = 0.289 \times 4.17 \times 10^{19} = 1.21 \times 10^{19} \, \text{decays/min} \approx 2.02 \times 10^{17} \, \text{Bq}\approx 5.45 \, \text{Ci} \]**At 1 hour:** \[ A(60) = 0 \, \text{decays/min} = 0 \, \text{Bq} = 0 \, \text{Ci} \]

Key Concepts

Half-life

The concept of half-life is crucial in understanding radioactive decay. It refers to the time it takes for half of the radioactive nuclei in a sample to decay. For a substance with a half-life, after one half-life period, only 50% of the original nuclei remain. In this exercise, the isotope ^{215} ext{Bi} has a half-life of 2.4 minutes. This means every 2.4 minutes, half of any present ^{215} ext{Bi} atoms decay into another element.

• After 2.4 minutes, 50% will have decayed.
• After 4.8 minutes, 25% will remain.
• After 7.2 minutes, 12.5% will remain.

To calculate how many undecayed nuclei remain after a certain period, we use the exponential decay formula: \[ N(t) = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}} \] where \( N_0 \) is the initial number of nuclei and \( t_{1/2} \) is the half-life.

Beta Decay

Beta decay is one type of radioactive decay where a beta particle (electron or positron) is emitted from an atomic nucleus. When a nucleus undergoes beta decay, a neutron is transformed into a proton or vice versa. For ^{215} ext{Bi}, this process involves a neutron becoming a proton, which results in the emission of an electron (beta particle) and an antineutrino. The nuclear equation for ^{215} ext{Bi} during beta decay is: \[^{215}_{83} \text{Bi} \rightarrow \, ^{215}_{84} \text{Po} \, + \, \beta^- \, + \, \bar{u}_{e} \] In this case, ^{215} ext{Bi} transforms into ^{215} ext{Po} (polonium-215). This transformation also increases the atomic number by one due to the increase of a proton in the nucleus while the mass number remains the same.

Nuclear Physics

Nuclear physics delves into the study of atomic nuclei and their interactions. The understanding of nuclear processes such as radioactive decay, nuclear fission, and fusion are fundamental to this field. In this context, beta decay, a core focus in nuclear physics, reveals how energy and particles are emitted from unstable atomic nuclei. It helps in comprehending why some nuclei transform and what governs the stability of atoms. This knowledge is crucial in multiple applications ranging from medicine to energy production. As we see in the exercise, beta decay provides a pathway for unstable ^{215} ext{Bi} to become more stable ^{215} ext{Po}. Each decay process is a balance between forces within the nucleus, primarily the weak nuclear force that facilitates beta decay.

Activity Measurement

The activity of a radioactive substance indicates how many decays occur per unit time, generally measured in becquerels (Bq) or curies (Ci). In nuclear physics, activity is a crucial measure since it tells us about the intensity of the radioactive decay process.

• A becquerel represents one decay per second.
• A curie, a larger unit, is equal to 3.7 \times 10^{10} Bq

To calculate the activity, we often use the decay constant \( \lambda \), which is related to the half-life \( t_{1/2} \) of the substance: \[ \lambda = \frac{\ln(2)}{t_{1/2}} \] Once \( \lambda \) is known, the activity \( A \) for a number of nuclei \( N \) is given by: \[ A = \lambda N \] At 10 minutes, for our ^{215} ext{Bi} sample, the calculation yields a significant activity indicating substantial decay, reducing drastically to zero by 60 minutes since nearly all nuclei have decayed.