Problem 101
Question
Half-life period of the radioactive element \(X\) is 10 hours. Amount of \(X\) left in the 11 th hour starting with one \(m o l X\) is (a) \((1 / 2)^{1 / 10}\) (b) \((1 / 2)^{11 / 10}\) (c) \((1 / 2)^{12 / 11}\) (d) \((1 / 2)^{1 / 11}\)
Step-by-Step Solution
Verified Answer
The amount of \(X\) left in the 11th hour is \((1 / 2)^{11 / 10}\).
1Step 1: Understanding the Half-Life Concept
The half-life of a radioactive element is the time it takes for half of the radioactive nuclei to decay. Here, the half-life of element \(X\) is 10 hours, meaning every 10 hours, half of the substance will decay.
2Step 2: Determine the Formula to Use
We use the formula for decay: \(N = N_0 \left( \frac{1}{2} \right)^{t/T}\), where \(N\) is the remaining quantity, \(N_0\) is the initial quantity, \(t\) is the time elapsed, and \(T\) is the half-life period.
3Step 3: Assign Values and Substitute
Given the problem, we have \(N_0 = 1\) mol, \(t = 11\) hours, and \(T = 10\) hours. Substitute these into the formula: \(N = 1 \cdot \left( \frac{1}{2} \right)^{11/10}\).
4Step 4: Simplify the Expression
Simplifying the expression \(N = \left( \frac{1}{2} \right)^{11/10}\), we find \(N = \left( \frac{1}{2} \right)^{11/10}\).
5Step 5: Select the Correct Answer Choice
Comparing our calculation \(\left( \frac{1}{2} \right)^{11/10}\) with the options given, the correct answer is option (b): \((1 / 2)^{11 / 10}\).
Key Concepts
Understanding Radioactive DecayExponential Decay FormulaHalf-Life of Radioactive Elements
Understanding Radioactive Decay
Radioactive decay is a natural process where an unstable atomic nucleus loses energy by emitting radiation. This process can transform an element into a different element or isotope over time.
Each radioactive isotope has a characteristic rate at which it decays. Key things to remember about radioactive decay include:
Each radioactive isotope has a characteristic rate at which it decays. Key things to remember about radioactive decay include:
- Radioactive decay is spontaneous and occurs without external influence.
- It results in the production of particles such as alpha, beta particles, or gamma rays.
- This decay is predictable on a statistical level yet random in individual events.
Exponential Decay Formula
In radioactive decay, the amount of material decreases exponentially over time. The decay is described by an exponential decay formula. This formula helps us calculate the remaining quantity of a material after a given time.
The formula is:
The formula is:
- \[N = N_0 \left( \frac{1}{2} \right)^{t/T}\]
- \(N\) is the remaining amount of the substance.
- \(N_0\) is the initial amount of the substance.
- \(t\) represents the time that has passed.
- \(T\) is the half-life, the time it takes for half of the material to decay.
Half-Life of Radioactive Elements
The half-life of a radioactive element is an essential concept in understanding how quickly a sample decays. Calculating the time it takes for half of the initial sample to transform into a different substance, half-life offers a time frame for decay processes.
Characteristics of half-life include:
Characteristics of half-life include:
- It is specific to each radioactive element or isotope and remains constant.
- Half-life is an indirect measure of the stability of a nucleus; stable isotopes have long half-lives, while unstable ones have short half-lives.
- It serves as a basis for techniques like radiocarbon dating and nuclear medicine.
Other exercises in this chapter
Problem 99
At radioactive equilibrium, the ratio between the atoms of two radioactive elements \(\mathrm{X}\) and \(\mathrm{Y}\) was found to be \(3.1 \times 10^{9}: 1\) r
View solution Problem 100
What weight of \(\mathrm{C}^{14}\) will have radioactivity one curie if \(\lambda\) (disintegration constant) is \(4.4 \times 10^{-12} \mathrm{sec}^{-1}\) ? (a)
View solution Problem 103
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
View solution Problem 104
In one type of mutual annihilation of an electron and a positron, three \gamma-ray photons are produced. If each photon has an energy of \(0.3407 \mathrm{MeV}\)
View solution