/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 129 A sample of radioactive substanc... [FREE SOLUTION] | 91Ó°ÊÓ

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A sample of radioactive substance is found \(90 \%\) of it's initial amount after one day. What \(\%\) of the original sample can be found after 3 days? (a) 81 (b) \(72.9\) (c) 25 (d) \(65.61\)

Short Answer

Expert verified
72.9% of the original sample can be found after 3 days.

Step by step solution

01

Understanding the Decay Process

The radioactive substance decays such that 90% of the initial amount remains after one day. This is a continuous decay process where the remaining amount can be found by multiplying the initial amount by 0.9 (which represents 90%) for each day that passes.
02

Calculating the Remaining Amount After 3 Days

To find the percentage of the original sample remaining after 3 days, raise 0.9 to the power of 3, because the decay happens each day and is independent. The formula used is: remaining percentage after n days = \( (0.9)^n \).
03

Computing and Expressing the Answer

For 3 days, the calculation is \( (0.9)^3 = 0.9 \times 0.9 \times 0.9 = 0.729 \). To express this as a percentage, multiply by 100, resulting in 72.9%.

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Key Concepts

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

Exponential Decay
Exponential decay denotes a process where a quantity decreases at a rate proportional to its current value. This concept is often encountered in radioactive decay problems, where the substance diminishes by a consistent percentage over equal time periods.

In our example, the substance decays to 90% of its initial amount after one day, which implies a decay rate of 10%. To mathematically represent this propensity to decrease over time, we make use of an exponential function.

The general formula for exponential decay is given by: \[ N(t) = N_0 \times (1 - r)^t \] where
  • \( N(t) \) is the amount at time t,
  • \( N_0 \) is the initial amount,
  • \( r \) is the decay rate (as a decimal),
  • \( t \) is the time elapsed.
In the context of our problem, with each passing day, we multiply the substance's remaining amount by 0.9, mirroring the concept's fundamentals.
Half-Life Calculation
The half-life of a radioactive substance is the time required for half of the substance to undergo decay. This period doesn't change, making it a unique characteristic of radioactive materials. Although our example does not directly ask for half-life, understanding this concept can help illustrate the decay process.

To determine half-life when given a decay percentage, one can use the formula: \[ t_{1/2} = \frac{\ln(2)}{\ln(1/(1 - r))} \] where
  • \( t_{1/2} \) is the half-life,
  • \( r \) is the decay rate.
Given our problem's 10% decay rate over one day, we can calculate the half-life and find it offers an alternative perspective to the same underlying exponential decay process.
Radioactive Substances
Radioactive substances are elements or isotopes that emit radiation as a result of nuclear instability. Most commonly, these substances undergo decay, releasing particles and energy, and transforming into different elements or isotopes over time.

Radioactivity has vast implications in fields such as medicine, archaeology (carbon dating), and energy (nuclear power generation). Radioactive decay is a random process at the level of single atoms, meaning that it is impossible to predict exactly when a particular atom will decay. However, for a large number of atoms, statistical trends emerge, allowing us to use exponential decay models like the one in our example to predict behavior of a radioactive sample over time.
Physical Chemistry Questions
Physical chemistry questions often delve into the behavior of substances on an atomic and molecular level, involving calculations related to kinetics, thermodynamics, and quantum mechanics. In the realm of kinetics, we study the rates of chemical reactions and processes like radioactive decay.

Problems like the one presented demonstrate how physical chemistry applies mathematical concepts to describe natural processes. By solving these exercises, students gain a deeper understanding of the laws governing the physical world. The ability to calculate the remaining amount of a radioactive substance, for instance, is crucial in fields such as environmental science, safety protocols, and medical treatments.

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Most popular questions from this chapter

For a reaction the initial rate is given as: \(R_{0}=k[A]_{0}^{2}[B]_{0}\) by what factor, the initial rate of reaction will increase if initial concentration of \(A\) is taken \(1.5\) times and of \(B\) is tripled? (a) \(4.5\) (b) \(2.25\) (c) \(6.75\) (d) None of these

\(\frac{k_{35^{\circ}}}{k_{34^{\circ}}}>1\), this means that (a) Rate increases with the rise in temperature (b) Rate decreases with rise in temperature (c) rate does not change with rise in temperature (d) None of the above

For the ideal gaseous reaction, the rate is generally expressed in terms of \(\frac{d P}{d t}\) instead of \(\frac{d C}{d t}\) or \(\frac{d n}{d t}\) (where \(C=\frac{n}{V}\) is concentration and \(n\) the no. of moles). What is the reaction among these three expressions if \(T\) and \(V\) are constant? (a) \(\frac{d C}{d t}=\frac{d n}{d t}=\frac{d P}{d t}\) (b) \(\frac{d C}{d t}=\frac{1}{V} \frac{d n}{d t}=\frac{1}{R T}\left(\frac{d P}{d t}\right)\) (c) \(R T \frac{d C}{d t}=\frac{d n}{d t}=\frac{d P}{d t}\) (d) None of these

The time of decay for a nuclear reaction is given by \(t=4 t_{1 / 2}\). The relation between the mean life \((T)\) and time of decay \((t)\) is given by : (a) \(2 T \ln 2\) (b) \(4 T \ln 2\) (c) \(2 T^{4} \ln 2\) (d) \(\frac{1}{T^{2}} \ln 2\)

The amount of \({ }_{6} \mathrm{C}^{14}\) isotope in a piece of wood is found to one fourth \((1 / 4)\) of that present in a fresh piece of wood. Calculate the age of the piece of wood \(\left(t_{12}\right.\) of \({ }_{6} \mathrm{C}^{14}=5770\) years) (a) 7999 year (b) 11540 year (c) 16320 year (d) \(23080 \mathrm{v}\)

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