Question

How long does it take for 2.00 g of niobium- 94 to decay to \(0.0625 \mathrm{~g}\) if its half-life is 20,000 y?

Short answer

It takes 100,000 years for niobium-94 to decay from 2.00 g to 0.0625 g.

Step-by-step solution

Understand the Problem

We need to find out how long it takes for a mass of niobium-94, initially at 2.00 g, to decay to 0.0625 g. We know the half-life is 20,000 years.

Use the Half-life Formula

The number of half-lives () it takes for a quantity to decay to a certain amount can be calculated using the formula:\[n = \frac{\log(\frac{\text{final amount}}{\text{initial amount}})}{\log(0.5)}\]

Substitute Known Values

Substitute the given values into the formula:\[n = \frac{\log(\frac{0.0625}{2.00})}{\log(0.5)}\]The calculation becomes:\[n = \frac{\log(0.03125)}{\log(0.5)}\]

Calculate the Number of Half-lives

Find the value of \(\log(0.03125)\) and \(\log(0.5)\):\[\log(0.03125) \approx -1.50515\]\[\log(0.5) \approx -0.30103\]Now, calculate:\[n = \frac{-1.50515}{-0.30103} \approx 5\]So, it takes approximately 5 half-lives for niobium-94 to decay from 2.00 g to 0.0625 g.

Calculate Time

Since each half-life takes 20,000 years, we multiply the number of half-lives by the duration of one half-life:\[\text{Time} = 5 \times 20,000 = 100,000 \text{ years}\]

Verification

Verify the calculation by checking if the decay happens as calculated. Each half-life reduces the mass by half: 1st half-life: 2.00 g → 1.00 g 2nd half-life: 1.00 g → 0.50 g 3rd half-life: 0.50 g → 0.25 g 4th half-life: 0.25 g → 0.125 g 5th half-life: 0.125 g → 0.0625 g The result matches; therefore, the solution is verified.

Key concepts

Decay Constant

Decay constant is a crucial concept in understanding the behavior of radioactive substances. It denotes the probability of a single atom decaying per unit time. This probability is an inherent property of each radioactive isotope.
To calculate the decay constant (\( \lambda \)), we often use the formula: \[\lambda = \frac{\ln(2)}{T_{1/2}}\] where \( T_{1/2} \) is the half-life of the substance. The constant provides insight into how quickly a nuclear decay process occurs. A higher decay constant indicates a faster decay process.
Knowing the decay constant helps us predict the behavior of radioactive materials over time. It is essential for many practical applications, such as dating archaeological finds or managing nuclear waste. Understanding the decay constant is fundamental when dealing with radioactive decay calculations.

Radioactive Decay

Radioactive decay is a process by which an unstable atomic nucleus loses energy by emitting radiation. This radiation may include alpha particles, beta particles, or gamma rays. This decay results in the transformation of the original nucleus into a stable one.
The decay process is spontaneous and statistically random, but it follows predictable patterns on a large scale. Every radioactive element has a characteristic half-life, which is the time it takes for half of the sample to decay. For niobium-94, the half-life is 20,000 years. This means if you start with 2.00 g, half of it will decay in 20,000 years.
This concept helps to understand the exponential nature of decay. As seen in the problem, every half-life significantly decreases the quantity of the substance remaining. Understanding this principle is key to correctly applying the half-life formula and solving decay problems.

Logarithmic Calculation

Logarithmic calculations are vital when handling exponential processes like radioactive decay. These calculations allow us to work with multiplication and division of large numbers more easily. In the exercise, we used logarithms to find how many half-lives it would take for 2.00 g of niobium-94 to decay to 0.0625 g.
We apply the formula \[n = \frac{\log(\frac{\text{final amount}}{\text{initial amount}})}{\log(0.5)}\]. This formula calculates the number of half-lives (n) needed for the decay. The logarithm simplifies the complex division of the amounts.
Understanding logarithmic calculations helps not only in radioactive decay problems but also in many scientific and financial calculations. Mastering this technique is essential for tackling problems involving exponential growth or decay, such as in populations, bank interests, or radioactivity.