Problem 103 - Introductory Chemistry Atoms First

Question

Polonium-210 is an alpha emitter and has a half-life of 138 days. (a) Write the equation for the radioactive decay of polonium-210. (b) How long will it take before only \(5.00 \%\) of the original amount of \({ }^{210}\) Po in a sample remains?

Step-by-Step Solution

Answer

a) The radioactive decay equation of polonium-210 is: \({}^{210}\text{Po} \rightarrow {}^{206}\text{Pb} + {}^{4}\text{He} + \gamma\). b) It takes approximately 598.5 days for only 5% of the original amount of polonium-210 in a sample to remain.

Step 1: a) Radioactive decay equation of polonium-210

The radioactive decay of polonium-210 (symbol: Po, atomic number: 84) can be represented as: \[ {}^{210}\text{Po} \rightarrow {}^{206}\text{Pb} + {}^{4}\text{He} + \gamma \] Here, polonium-210 (Po) decays to lead-206 (Pb), releasing an alpha particle (He) and gamma radiation (γ).

Step 2: b) Calculate the decay constant

To determine how long it will take for 5% of the original polonium-210 to remain, first, we need to calculate the decay constant (λ). The half-life (T) of polonium-210 is given as 138 days. The decay constant can be determined using the equation: \[ T = \frac{\ln{2}}{\lambda} \] Rearrange the equation to solve for λ: \[ \lambda = \frac{\ln{2}}{T} \] Now, we can plug in the half-life of polonium-210 (T = 138 days) and calculate λ: \[ \lambda = \frac{\ln{2}}{138} \approx 0.00502 \: \text{days}^{-1} \]

Step 3: b) Determine the time for 5% of the original amount to remain

To find how long it takes for only 5% of the original amount of polonium-210 to remain, we can use the exponential decay model: \[ N(t) = N_{0}e^{-\lambda t} \] Where: \(N(t)\) is the amount of polonium-210 remaining after time t, \(N_{0}\) is the initial amount of polonium-210, and \(e\) is the base of the natural logarithm (approximately equal to 2.718). First, we will express the percentage of remaining polonium-210 in terms of the initial amount: \[ N(t) = 0.05N_{0} \] Next, we will plug this expression into our decay equation, isolating the variable t: \[ 0.05N_{0} = N_{0}e^{-\lambda t} \] Now, we can divide both sides by \(N_{0}\): \[ 0.05 = e^{-\lambda t} \] Take the natural logarithm of both sides: \[ \ln{0.05} = -\lambda t \] Finally, isolate t and plug in the value of λ we calculated earlier: \[ t = \frac{\ln{0.05}}{-0.00502 \: \text{days}^{-1}} \approx 598.5 \: \text{days} \] So, it takes approximately 598.5 days for only 5% of the original amount of polonium-210 in a sample to remain.

Key Concepts

Alpha Emission

Alpha emission is a type of radioactive decay where an unstable atom releases an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, essentially a helium nucleus. This process helps the original atom achieve a more stable state.
In the case of polonium-210, alpha emission results in the transformation of polonium into lead-206.

• Equation: \( {}^{210}\text{Po} \rightarrow {}^{206}\text{Pb} + {}^{4}\text{He} + \gamma \)
• An alpha particle \( {}^{4}\text{He} \) is released.
• Gamma radiation \( \gamma \) may also be emitted, though it's not always shown.

This decay process decreases the atomic number by 2 and the mass number by 4, changing the element's identity.

Half-Life Calculation

The half-life of a radioactive substance is the time it takes for half of the material to decay. For polonium-210, this time is 138 days. Understanding the half-life is crucial because it helps us predict how long a radioactive material remains active.
Calculating half-life allows scientists and engineers to determine how long it will take for the substance to reduce to a more manageable level. It's especially important in radioactive waste management and medical applications where timing is critical.

• Key Formula: \( T = \frac{\ln{2}}{\lambda} \)
• Given: For polonium-210, \( T = 138 \) days.

Knowing the half-life helps calculate the decay constant and other important parameters in radioactive decay studies.

Exponential Decay Model

The exponential decay model describes how the quantity of a radioactive substance decreases over time. This model follows an exponential curve, meaning the decay rate is proportional to the current amount of substance.
In our example, the decay of polonium-210 follows this model, which can be represented mathematically. It helps us comprehend how a substance decays over both short and long periods.

• Equation: \( N(t) = N_{0}e^{-\lambda t} \)
• Where:
  • \( N(t) \) is the amount remaining after time \( t \).
  • \( N_{0} \) is the initial amount.
  • \( \lambda \) is the decay constant.

This model provides a clear picture of how rapidly or slowly a radioactive material loses its initial mass.

Decay Constant

The decay constant, \( \lambda \), represents the probability per unit time that a nucleus will decay. It's a key factor in determining the speed at which a radioactive substance decays.
For polonium-210, the decay constant was calculated using its known half-life.

• Formula: \( \lambda = \frac{\ln{2}}{T} \)
• Calculation: Given \( T = 138 \) days, \( \lambda = \frac{\ln{2}}{138} \approx 0.00502 \text{ days}^{-1} \)

The decay constant lets us use the exponential decay model effectively to find out how much substance remains after a certain amount of time. It's crucial for calculations involving both short-lived and long-lived isotopes.