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If the amount of radioactive substance is increases three times, the number of disintegration per unit time will be (a) doubled (b) one-third (c) triple (d) uncharged

Short Answer

Expert verified
The number of disintegrations per unit time will be tripled.

Step by step solution

01

Identify the Law governing radioactive decay

The number of disintegrations per unit time, often referred to as the activity (A), of a radioactive substance is directly proportional to the amount of the substance present. This is described by the radioactive decay law which states that A is proportional to N, where N is the number of radioactive atoms. Mathematically, this can be expressed as A = λN, where λ is the decay constant.
02

Analyze the change in the amount of radioactive substance

If the amount of the radioactive substance is increased three times, then the new number of radioactive atoms, labeled N', would be three times the initial amount, N. Hence, N' = 3N.
03

Apply the decay law to find the new activity

Using the proportional relationship, the new activity A' can be found by substituting N' into the decay law equation. Hence, A' = λN' = λ(3N) = 3λN. Since λN is the initial activity A, the new activity A' is three times the initial activity, A' = 3A.

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Key Concepts

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

Radioactive Substance
A radioactive substance is a material that contains unstable atoms which release energy in the form of radiation as they decay to become more stable. This process of decay occurs spontaneously and at a constant rate, unique to each radioactive isotope. The decay can result in the emission of alpha particles, beta particles, gamma rays, and other forms of radiation.

Each radioactive substance has a characteristic half-life, the time it takes for half of the radioactive atoms to decay. This property is fundamental to identifying the type of radioactive substance and predicting how its activity will change over time. In the context of the textbook exercise, we consider the relationship between the amount of radioactive substance present and the number of disintegrations per unit time, which corresponds to the substance's activity.
Disintegrations Per Unit Time
The term disintegrations per unit time refers to the number of radioactive decays occurring in a substance per unit time, often measured in seconds. This measure is also known as the activity of the radioactive substance. The activity is a crucial parameter because it represents the rate at which the substance undergoes radioactive decay.

By monitoring the disintegrations per unit time, scientists can deduce valuable information about the sample's current state and its future behavior. In our exercise, the direct proportionality between the amount of substance and its activity leads to a straightforward conclusion: tripling the amount of a radioactive substance results in tripling the activity, meaning the disintegrations per unit time also triple.
Decay Constant
The decay constant, symbolized by λ, represents the probability per unit time of a single radioactive nucleus decaying. It is a fundamental property of each radioactive isotope and plays a pivotal role in describing the kinetics of radioactive decay. The inverse of the decay constant is the mean lifetime of the atoms, which provides the average time an atom will exist before it decays.

The radioactive decay law is mathematically expressed as A = λN, where A denotes the activity and N the number of undecayed atoms. As evidenced by the exercise solution, when the amount of radioactive substance is increased, the decay constant remains unchanged. This is because it is an intrinsic property of the substance, not affected by the quantity present. Therefore, when the amount of substance increases threefold, the number of disintegrations per unit time is also tripled, as the decay rate per atom stays constant.

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Most popular questions from this chapter

Consider the following process of decay, \({ }_{92} \mathrm{U}^{234} \rightarrow{ }_{90} \mathrm{Th}^{230}+{ }_{2} \mathrm{He}^{4} ; t_{1 / 2}=2,50,000\) years \({ }_{90} \mathrm{Th}^{230} \rightarrow{ }_{88} \mathrm{Ra}^{226}+{ }_{2} \mathrm{He}^{4} ; t_{1 / 2}=80,000\) years \({ }_{88} \mathrm{Ra}^{226} \rightarrow{ }_{86} \mathrm{Rn}^{222}+{ }_{2} \mathrm{He}^{4} ; t_{1 / 2}=1600\) years After the above process has occurred for a long time, a state is reached where for every two thorium atoms formed from \({ }_{92} \mathrm{U}^{234}\), one decomposes to form \({ }_{88} \mathrm{Ra}^{226}\) and for every two \({ }_{88} \mathrm{Ra}^{226}\) formed, one decomposes. The ratio of \({ }_{90} \mathrm{Th}^{230}\) to \({ }_{88} \mathrm{Ra}^{226}\) will be (a) \(250000 / 80000\) (b) \(80000 / 1600\) (c) \(250000 / 1600\) (d) \(251600 / 8\)

In \(\beta\) -decay, an electron comes out from an atom. The electron comes out due to nuclear change, not from the orbit of the atom. It may be explained by the fact that on \(\beta\) -decay, (a) the atomic number increases by one unit. (b) the mass number remains unchanged. (c) the atomic species get changed. (d) the atomic species remains unchanged.

Consider the beta decay, \(\mathrm{Au}^{198} \rightarrow \mathrm{Hg}^{198^{*}}\), where \(\mathrm{Hg}^{198^{*}}\) represents a mercury nucleus in an excited state at energy \(1.063 \mathrm{MeV}\) above the ground state. What can be the maximum kinetic energy of the electron emitted? The atomic masses of \(\mathrm{Au}^{198}\) and \(\mathrm{Hg}^{198}\) are \(197.968 \mathrm{u}\) and \(197.966 \mathrm{u}\), respec- tively. \((1 \mathrm{u}=931.5 \mathrm{MeV})\) (a) \(0.8 \mathrm{MeV}\) (b) \(1.863 \mathrm{MeV}\) (c) \(1.063 \mathrm{MeV}\) (d) \(1.0 \mathrm{MeV}\)

The activity of a certain preparation decreases \(2.5\) times after \(7.0\) days. Find its half-life. (a) \(10.58\) days (b) \(2.65\) days (c) \(5.3\) days (d) \(4.2\) days

A radioactive sample has an initial activity of 28 dpm. Half hour later, the activity is \(14 \mathrm{dpm}\). How many atoms of the radioactive nuclide were there originally? \((\ln 2=0.7)\) (a) 1200 (b) 200 (c) 600 (d) 300

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