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Two radioactive elements A and B have decay constant \(\lambda\) and \(10 \lambda\) respectively. If the decay begins with the same number of atoms of the \(\mathrm{n}\), the ratio of atoms of \(\mathrm{A}\) to those of B after time \(1 / 9 \lambda\) will be (a) \(\mathrm{e}^{-3}\) (b) \(\mathrm{e}^{2}\) (c) \(\mathrm{e}\) (d) \(\mathrm{e}^{-1}\)

Short Answer

Expert verified
The ratio of atoms of A to B is \( e \), so the answer is (c) \( e \).

Step by step solution

01

Understanding Decay Law

The radioactive decay law states that the number of atoms decaying per unit time is proportional to the current number of atoms. Mathematically, this is represented as \( N(t) = N_0 e^{-\lambda t} \), where \( N(t) \) is the number of atoms at time \( t \), \( N_0 \) is the initial number of atoms, and \( \lambda \) is the decay constant.
02

Apply Decay Law to Element A

For element A, the decay constant is \( \lambda \). Using the decay law, the number of atoms of element A at time \( t = \frac{1}{9\lambda} \) is \( N_A = N_0 e^{-\lambda \cdot \frac{1}{9\lambda}} = N_0 e^{- rac{1}{9}} \).
03

Apply Decay Law to Element B

For element B, the decay constant is \( 10\lambda \). Similarly, the number of atoms of element B at time \( t = \frac{1}{9\lambda} \) is \( N_B = N_0 e^{-10\lambda \cdot \frac{1}{9\lambda}} = N_0 e^{- rac{10}{9}} \).
04

Calculate the Ratio

To find the ratio of atoms of A to atoms of B after the given time, divide the expressions obtained in Step 2 and Step 3. The ratio is \( \frac{N_A}{N_B} = \frac{N_0 e^{- rac{1}{9}}}{N_0 e^{- rac{10}{9}}} = e^{- rac{1}{9} + \frac{10}{9}} = e^{\frac{9}{9}} = e \).

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Key Concepts

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

Decay Constant
In radioactive decay, a vital concept is the decay constant, symbolized as \( \lambda \). It expresses how quickly a radioactive substance transforms into another element or isotope. The decay constant is the proportionality factor in the formula that represents how many atoms disintegrate per unit time. It reflects the intrinsic ability of the substance to undergo decay and is unique to each radioisotope.

For example, if a substance has a high decay constant, it means it decays rapidly. Conversely, a low decay constant indicates slower decay.

This principle can be easily viewed mathematically in the radioactive decay law, where the decay constant directly influences the exponential function that represents the decay process. Understanding the decay constant is crucial for predicting how much of a radioisotope will remain after a certain period.
Radioactive Decay Law
The radioactive decay law formulates the process of radioactive transformation mathematically and helps us predict the remaining quantity of a substance over time. The law is represented by the equation: \( N(t) = N_0 e^{-\lambda t} \).

Here:
  • \(N(t)\) is the number of atoms left at time \(t\).
  • \(N_0\) is the initial quantity of the substance at time zero.
  • \(\lambda\) is the decay constant.
  • \(t\) is the time elapsed.
This equation shows us that the number of atoms diminishes exponentially with time. As time increases, \(N(t)\) reduces at an exponential rate, determined by \(\lambda\).

In essence, the radioactive decay law gives a comprehensive view of how and at what rate a radioactive substance breaks down over a period, dictating how quickly it becomes stable or transforms into another isotope.
Atoms Ratio Calculation
Calculating the ratio of the number of atoms after a given time between two radioactive substances involves using the decay law for each substance and comparing them. For two substances, A and B, each with different decay constants, it becomes necessary to input their values into the decay equation.

Suppose we take two elements, A and B, with decay constants \(\lambda\) and \(10\lambda\) respectively. After a specific time, \(t\), the number of remaining atoms can be given as:
  • For A: \(N_A = N_0 e^{-\lambda t}\)
  • For B: \(N_B = N_0 e^{-10\lambda t}\)
To find the ratio \(\frac{N_A}{N_B}\), we divide these expressions:

\[\frac{N_A}{N_B} = \frac{N_0 e^{-\lambda t}}{N_0 e^{-10\lambda t}} = e^{-\lambda t + 10\lambda t} = e^{9\lambda t}\]

This reveals how the comparative rate of decay affects their proportions over time. Such calculations are fundamental in nuclear physics for monitoring isotopic changes and predicting future concentrations in decay chains.

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Most popular questions from this chapter

One microgram of radioactive sodium \({ }_{11} \mathrm{Na}^{24}\) with a half- life of 15 hours was injected into a living system for a bio assay. How long will it take for the radioactivity to fall to \(25 \%\) of the initial value? (a) 60 hours (b) \(22.5\) hours (c) 375 hours (d) 30 hours

A radioactive substance having a half-life of 3 days was received in 12 days. It was found that there was only \(3 \mathrm{~g}\) of the isotope in the container. The initial weight of the isotope when packed was (a) \(12 \mathrm{~g}\) (b) \(24 \mathrm{~g}\) (c) \(48 \mathrm{~g}\) (d) \(96 \mathrm{~g}\)

Lead is the final product formed by a series of changes in which the rate determining stage is the radioactive decay of uranium-238. This radioactive decay is a first order reaction with a half-life of \(4.5 \times 10^{9}\) years. What would be the age of a rock sample originally lead free, in which the molar proportion of uranium to lead is now \(1: 3\) ? (a) \(1.5 \times 10^{9}\) years (b) \(2.25 \times 10^{9}\) years (c) \(4.5 \times 10^{9}\) years (d) \(9.0 \times 10^{9}\) years

In natural radioactive disintegration, U-238 emits one \(\alpha\) and two \(\beta\) and then five \(\alpha\) particle successively. The end product obtained is (a) \({ }_{82} \mathrm{~Pb}^{218}\) (b) \({ }_{82} \mathrm{~Pb}^{214}\) (c) \({ }_{84} \mathrm{~Pb}^{218}\) (d) \({ }_{82} \mathrm{~Pb}^{216}\)

Atom bomb is based on the principle of (a) nuclear fusion (b) nuclear fission (c) nuclear transformation (d) carbon dating

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