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Half-life period of the radioactive element \(X\) is 10 hours. Amount of \(X\) left in the 11 th hour starting with one \(\mathrm{mol} \mathrm{X}\) is (a) \((1 / 2)^{1 / 10}\) (b) \((1 / 2)^{11 / 10}\) (c) \((1 / 2)^{12 / 11}\) (d) \((1 / 2)^{1 / 11}\)

Short Answer

Expert verified
The amount of \(X\) left in the 11th hour is \((1/2)^{11/10}\). Option (b).

Step by step solution

01

Understand the half-life concept

The half-life of a radioactive substance is the time taken for half of the original amount of the substance to decay. If the half-life of element \(X\) is 10 hours, it means that after every 10 hours, only half of the original amount of \(X\) will remain.
02

Set up the decay formula

The decay of a radioactive element can be described by the equation: \[ N(t) = N_0 \times \left(\frac{1}{2}\right)^{t/T_{1/2}} \] where, - \(N(t)\) is the remaining quantity after time \(t\), - \(N_0\) is the initial quantity,- \(T_{1/2}\) is the half-life period, - \(t\) is the time elapsed.
03

Substitute values into the formula

Given the initial amount \(N_0 = 1\) mol, half-life \(T_{1/2} = 10\) hours, and we want to find the amount left after \(t = 11\) hours. Substitute into the formula:\[ N(11) = 1 \times \left(\frac{1}{2}\right)^{11/10} \]
04

Simplify the expression

Calculating a power involving fractions, the expression \(\left(\frac{1}{2}\right)^{11/10}\) simplifies as it is. Thus, \(N(11) = \left(\frac{1}{2}\right)^{11/10}\).

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Key Concepts

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

Half-Life
The concept of half-life is important for understanding how radioactive decay works. Half-life is the time required for half of a radioactive substance to decay. For example, if the half-life of an element is 10 hours, after 10 hours, only half of the original amount of the element remains. Another 10 hours later, only a quarter (half of the remaining half) would be left.

In simple terms, if you start with a full amount of a radioactive material, over each half-life period, you'll have half of what you started with. This is a consistent process, meaning that no matter how much time has passed, at the end of each half-life, the remaining amount is halved again.
  • If you start with 1 mol, after 1 half-life, you have 0.5 mol.
  • After 2 half-lives, you have 0.25 mol.
  • And after 3 half-lives, you have 0.125 mol, and so on.
This makes it a predictable and quantifiable way to understand how substances decay over time.
Decay Formula
The decay formula for radioactive materials allows us to calculate the remaining quantity of the substance after a given amount of time. It uses the relation involving half-life and the time elapsed. The standard formula is:

\[ N(t) = N_0 \times \left(\frac{1}{2}\right)^{t/T_{1/2}} \]

Here, \(N(t)\) is the amount left after time \(t\). \(N_0\) is the original amount of substance you started with. The \(T_{1/2}\) represents the half-life of the material, indicating the interval at which the initial quantity gets halved.

  • This formula shows that the remaining amount of a radioactive material lessens exponentially over time.
  • The fraction \(\left(\frac{1}{2}\right)^{t/T_{1/2}}\) conveys the continual halving process; \(t\) over \(T_{1/2}\) tells us how many half-lives have passed.
In practice, by substituting known values into this formula, you can compute how much of the substance is left after any given period.
Quantities and Units
In the context of radioactive decay, it's crucial to understand the quantities and units involved to precisely express and compute how substances change over time.

When dealing with the decay of an element, like in the exercise, we use mol as a unit for the substance's amount. A mol is a standard unit in chemistry representing a specific number of particles (usually atoms or molecules) of a substance.

For half-life \(T_{1/2}\), it's typically measured in units of time such as seconds, minutes, or hours depending on how fast the decay occurs. This is essential for consistency and accuracy in calculations.
  • \(N_0\) is expressed in moles or similar units indicating amount.
  • The time \(t\), during which decay is observed, should have the same unit as the half-life \(T_{1/2}\) to maintain the equation’s integrity.
Understanding these quantities and ensuring unit consistency allows accurate predictions of how much radioactive material remains after any given period.

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Most popular questions from this chapter

A sample of \(\mathrm{U}^{238}\left(\mathrm{t}_{1 / 2}=4.5 \times 10^{9} \mathrm{yrs}\right)\) ore is found con- taining \(23.8 \mathrm{~g} \mathrm{U}^{238}\) and \(20.6 \mathrm{~g}\) of \(\mathrm{Pb}^{206} .\) Calculate the age of the ore. (a) \(4.9 \times 10^{9}\) year (b) \(9.0 \times 10^{11}\) year (c) \(9.4 \times 10^{9}\) year (d) \(4.5 \times 10^{9}\) year

A sample of radioactive substance gave 630 counts per minute and 610 counts per minute at times differing by 1 hour. The decay constant \((\lambda)\) in \(\min ^{-1}\) is given by (a) \(\lambda=\frac{630}{610} \times 60\) (b) \(\mathrm{e}^{60 \mathrm{k}}=\frac{630}{610}\) (c) \(\lambda=\frac{2.303}{60} \log \frac{610}{630}\) (d) \(\lambda=\frac{2.303}{60} \times \frac{630}{610}\)

Two radioactive elements \(\mathrm{A}\) and \(\mathrm{B}\) 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 \(\mathrm{B}\) after time \(1 / 9 \lambda\) will be (a) \(\mathrm{e}^{-3}\) (b) \(\mathrm{e}^{2}\) (c) \(\mathrm{e}\) (d) \(\mathrm{e}^{-1}\)

One of the hazards of nuclear explosion is the generation of \(\mathrm{Sr}^{90}\) and its subsequent incorporation in bones. This nuclide has a half life of \(28.1\) years. Suppose one microgram was absorbed by a new born child, how much \(\mathrm{Sr}^{90}\) will remain in his bones after 20 years? (a) \(61 \mu \mathrm{g}\) (b) \(61 \mathrm{~g}\) (c) \(0.61 \mu \mathrm{g}\) (d) none

The half-life of a radio isotope is four hours. If the initial mass of the isotope was \(200 \mathrm{~g}\) the mass remaining undecayed after 24 hours is (a) \(2.084 \mathrm{~g}\) (b) \(3.125 \mathrm{~g}\) (c) \(4.167 \mathrm{~g}\) (d) \(1.042 \mathrm{~g}\)

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