Question

How much time is required for a \(6.25-\mathrm{mg}\) sample of \({ }^{51} \mathrm{Cr}\) to decay to \(0.75 \mathrm{mg}\) if it has a half-life of \(27.8\) days?

Short answer

It takes approximately 87.4 days for a 6.25 mg sample of Chromium-51 to decay to 0.75 mg, given its half-life of 27.8 days.

Step-by-step solution

Label the known and unknown values

The initial mass of the sample (\(N_0\)) is 6.25 mg, the remaining mass of the sample after time \(t\) (\(N(t)\)) is 0.75 mg, and the half-life (\(T\)) is 27.8 days. We want to find the time (\(t\)) it takes for the sample to decay from 6.25 mg to 0.75 mg.

Prepare the decay formula

Write the decay formula as: \(N(t) = N_0 \cdot (\frac{1}{2})^\frac{t}{T}\)

Fill in the given values

Now plug in the values of \(N(t)\), \(N_0\), and \(T\) in the decay formula: \(0.75 = 6.25 \cdot (\frac{1}{2})^\frac{t}{27.8}\)

Solve for t

In order to solve for \(t\), we first divide both sides by 6.25: \(\frac{0.75}{6.25} = (\frac{1}{2})^\frac{t}{27.8}\) Now take the logarithm (base 2) of both sides of the equation: \(log_2{(\frac{0.75}{6.25})} = log_2{(\frac{1}{2})^\frac{t}{27.8}}\) Apply the logarithm property \(\log_b{(a^c)} = c\log_b{a}\): \(log_2{(\frac{0.75}{6.25})} = \frac{t}{27.8} \cdot log_2{(\frac{1}{2})}\) Now, divide both sides by \(log_2{(\frac{1}{2})}\): \(\frac{log_2{(\frac{0.75}{6.25})}}{log_2{(\frac{1}{2})}} = \frac{t}{27.8}\) Then multiply both sides by 27.8 to get \(t\): \(t = 27.8\cdot\frac{log_2{(\frac{0.75}{6.25})}}{log_2{(\frac{1}{2})}}\)

Calculate the value for t

Now we can calculate the value for \(t\): \(t = 27.8\cdot\frac{log_2{(\frac{0.75}{6.25})}}{log_2{(\frac{1}{2})}} \approx 87.399\) So it takes about 87.4 days for the 6.25 mg sample of Chromium-51 to decay to 0.75 mg.

Key concepts

Half-life

In the realm of chemistry and physics, the concept of half-life is fundamental, especially when dealing with radioactive substances. The term "half-life" refers to the amount of time it takes for half of a sample of a radioactive element to decay. This does not mean the element disappears entirely; instead, it undergoes a transformation into another element or isotope.
For any radioactive element, the half-life is constant and unique, making it a crucial characteristic for predicting radioactive decay. When a scientist knows the half-life of a substance, they can determine how much of the substance will remain after any given period.

• A shorter half-life signifies a faster decay rate.
• A longer half-life implies slower decay.

This consistent rate of decay simplifies complex calculations into more manageable predictions, allowing chemists and physicists to date objects and calculate the time required for a substance to decay to a specific level, just as with Chromium-51 in the given problem.

Chromium-51

Chromium-51 is an isotope of chromium commonly used in various scientific and medical applications. Due to its radioactive nature, it is especially useful in studies involving blood cells and biological processes.
This isotope is favored because of its relatively short half-life of approximately 27.8 days, which allows it to decay to safer levels within a reasonable timeframe after usage.
In medical diagnostics, Chromium-51 often assists in tracing blood volume and survival studies of red blood cells. Because of its distinct half-life, scientists can precisely calculate decay and application rates, ensuring controlled exposure during medical and research procedures.

Exponential Decay Formula

An essential mathematical tool when working with radioactive decay is the exponential decay formula. This formula quantifies how the quantity of a substance decreases over time due to decay. Specifically, it can be represented as:
\[ N(t) = N_0 \cdot \left( \frac{1}{2} \right)^{\frac{t}{T}} \]

• \(N(t)\) is the remaining quantity of the substance after time \(t\).
• \(N_0\) represents the initial quantity of the substance.
• \(T\) stands for the half-life of the substance.

This formula shows that the quantity of the substance is reduced by half every half-life interval. It is derived from the primary characteristic of exponential functions where a constant is multiplied repeatedly.
This formula allows for precise calculations and predictions regarding how long it will take for a certain amount of a substance to decay to a specific level, as seen through the calculated half-life of Chromium-51.

Logarithms in Chemistry

Logarithms might seem more aligned with mathematics, but they play a critical role in chemistry, particularly when calculating decay and growth processes. They help simplify the exponential scale to a more manageable linear form.
In chemistry, when solving exponential decay equations, logarithms help unravel the unknown time period (\(t\)) required for decay. They convert multiplicative relationships into additive ones, allowing for easier manipulation of formulae, especially when solving for unknowns.

• For instance, when solving an equation such as \(\frac{0.75}{6.25} = \left(\frac{1}{2}\right)^{\frac{t}{27.8}}\), taking the logarithm on both sides simplifies it significantly.
• This step converts the problem to solve for \(t\) into a linear form, making computation more direct and comprehensible.

As such, understanding and utilizing logarithms in radioactive decay contexts is crucial for precisely determining decay times of isotopes like Chromium-51.