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Chapter 10: Problem 9

Bismuth- 210 has a half-life of 5 days. What is the decay rate per day?

Short Answer

Expert verified
Answer: The approximate decay rate per day for Bismuth-210 is 5%.

Step by step solution

01

Derive the decay rate constant (k) formula from the exponential decay formula

We start with the exponential decay formula: N(t) = N0 * (1/2)^(t/T) Now, we want to rewrite it to express the decay rate constant (k). By taking the natural logarithm of both sides of the equation, we have: ln(N(t)/N0) = ln((1/2)^(t/T)) We know that, ln(a^b) = b * ln(a), so: ln(N(t)/N0) = (t/T) * ln(1/2) Rearranging for the decay rate constant (k = t/T), we have: k = ln(N(t)/N0) / (t * ln(1/2))
02

Calculate the decay rate constant (k) using the given half-life

For Bismuth-210, the half-life T is given as 5 days. We now plug this value into the k formula, and then we calculate k when the decay rate reaches 50% (which is the point in time when half of the substance has decayed). k = ln(1/2) / (5 * ln(1/2)) k 鈮� 1/5
03

Calculate the decay rate per day

To find the decay rate per day, we can simply take the inverse of k: Decay Rate per Day = 1/k Decay Rate per Day 鈮� 1/(1/5) This simplifies to: Decay Rate per Day 鈮� 5 So, the decay rate of Bismuth-210 is approximately 5% per day.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Life
Half-life is a crucial concept in understanding exponential decay, particularly in radioactive substances. The half-life of a substance is the time required for half of its material to decay. Take an example like Bismuth-210. It has a half-life of 5 days, meaning every 5 days, only half of the Bismuth-210 remains from its original quantity.
The half-life helps us predict how long a substance will last or how fast it transforms. In an equation, the half-life is represented as 'T,' and it corresponds to the time taken for the substance to reduce to half its initial amount. By knowing T, we can calculate various decay properties like the decay rate constant.
  • The shorter the half-life, the faster the rate of decay.

  • The longer the half-life, the slower the decay process.
Decay Rate Constant
The decay rate constant (k) is a measure that determines how quickly a substance undergoes decay. For substances experiencing exponential decay, the decay rate constant offers insights into the rate at which the process occurs. Understanding k is essential for calculating the decay over various time periods.
To find the decay rate constant, we typically utilize the natural logarithm of the half-life equation. For any given half-life, we can rearrange the exponential decay formula to isolate k.
  • k is determined with the formula \( k = \frac{\ln(1/2)}{T} \), where T is the half-life.

  • In cases like Bismuth-210, with a known half-life of 5 days, this calculation results in a k value which reflects the rapidity of the decay process.
Natural Logarithm
The natural logarithm, often symbolized as ln, plays a pivotal role in solving exponential decay problems. Exponential decay equations often require logarithmic transformation to linearize them, making calculations easier.
The natural logarithm specifically makes use of the base \(e\), which is approximately equal to 2.718. By applying ln to an exponential equation, we can effectively dissect and understand the decay in terms of rates and time.
  • It is used to simplify equations such as \( \ln(N(t)/N_0) = \frac{t}{T} \cdot \ln(1/2) \), where N(t) is the remaining quantity after time t, and \(N_0\) is the initial quantity.

  • The transformation helps to derive values for decay rate constants and understand changes over regular time intervals.

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Most popular questions from this chapter

Solve the equations in Problem using the following approximations: $$ 10^{0.301}=2, \quad 10^{0.477}=3, \quad 10^{0.699}=5 . $$ Example. Solve \(10^{x}=6 .\) Solution. We have $$ \begin{aligned} 10^{x} &=6 \\ &=2 \cdot 3 \\ &=10^{0.301} \cdot 10^{0.477} \\ &=10^{0.301+0.477} \\ &=10^{0.778} \\ \text { so } x=0.778 . \end{aligned} $$ \(10^{x}=15\)

For the functions in the form \(P=a b^{t / T}\) describing population growth. (a) Give the values of the constants \(a, b\), and \(T\). What do these constants tell you about population growth? (b) Give the annual growth rate. \(P=400\left(\frac{2}{3}\right)^{t / 14}\)

In Exercises \(7-10\) give the growth factor that corresponds to the given growth rate. \(8.5 \%\) growth

The average rainfall in Hong Kong in January and February is about 1 inch each month. From March to June, however, average rainfall in each month is double the average rainfall of the previous month. (a) Make a table showing average rainfall for each month from January to June. (b) Write a formula for the average rainfall in month \(n,\) where \(2 \leq n \leq 6\) and January is month 1 . (c) What is the total average rainfall in the first six months of the year?

The balance in dollars in a bank account after \(t\) years is given by the function \(f(t)=4622(1.04)^{t}\) rounded to the nearest dollar. For which values of \(t\) in Table 10.10 is (a) \(f(t)<5199 ?\) (b) \(f(t)>5199 ?\) (c) \(f(t)=5199 ?\) What do your answers tell you about the bank account? $$ \begin{array}{c|c|c|c|c|c} \hline t & 0 & 1 & 2 & 3 & 4 \\ \hline 4622(1.04)^{t} & 4622 & 4807 & 4999 & 5199 & 5407 \\ \hline \end{array} $$
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