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Chapter 27: Problem 79

The half-life period of radium is \(1600 \mathrm{yr}\). The fraction of a sample of radium that would remain after \(6400 \mathrm{yr}\) is (a) \(\frac{1}{4}\) (b) \(\frac{1}{2}\) (c) \(\frac{1}{8}\) (d) \(\frac{1}{16}\)

Short Answer

Expert verified
(d) \(\frac{1}{16}\)

Step by step solution

01

Understand Half-life Concept

The half-life of a substance is the time required for half of the substance to decay. For radium, this is given as 1600 years. This means that every 1600 years, half of the sample remains.
02

Determine Number of Half-lives

Calculate how many half-life periods fit into 6400 years. Since one half-life period is 1600 years, divide 6400 by 1600:\[\text{Number of half-lives} = \frac{6400}{1600} = 4\]
03

Calculate Remaining Fraction After 4 Half-lives

After each half-life, the remaining fraction of the sample is halved. Start with 1 (the whole sample) and halve it four times:1. After 1 half-life: \(\frac{1}{2}\)2. After 2 half-lives: \(\frac{1}{4}\)3. After 3 half-lives: \(\frac{1}{8}\)4. After 4 half-lives: \(\frac{1}{16}\)Thus, after 6400 years, \(\frac{1}{16}\) of the original sample remains.

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Key Concepts

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

Radioactive Decay
Radioactive decay is a natural process by which an unstable atomic nucleus loses energy. This happens through the emission of particles or electromagnetic waves, transforming the original nucleus into a different element or a different isotope of the same element. Decay occurs at an unpredictable time for a single atom, but when dealing with a large number of atoms, the decay rate becomes a predictable statistical process.

This process is crucial for understanding how materials change over time, especially when dealing with radioactive substances like radium. The decay of such materials can be measured in terms of half-life. The half-life is the time it takes for half of a radioactive sample to decay into other elements or isotopes.
  • Random Process: Individual atoms decay without a set time, but collectively decay at a predictable rate.
  • Transformation: Results in a change to a different element or isotope.
  • Regulated by Half-life: Allows calculation of how much of a substance remains over time.
Radium
Radium is a highly radioactive element discovered by Marie and Pierre Curie in the late 19th century. Known for its intense radioactivity, radium emits alpha particles during its decay and eventually transforms into radon gas. Radium has historical significance due to its early use in luminous paints and cancer therapies.

Understanding the half-life of radium is essential in applications where knowing the duration of radioactivity is critical. With a half-life of 1600 years, radium decays quite slowly compared to other radioactive elements. This long half-life means that radium remains active and hazardous for extended periods.
  • Discovered By: Marie and Pierre Curie.
  • Main Characteristics: Emits alpha particles, transforming into radon.
  • Half-life: 1600 years, indicating slow decay and long-term radioactivity.
  • Applications: Used in historical luminous paints and as a cancer treatment.
Nuclear Physics
Nuclear physics is a branch of physics that studies the components and forces within atomic nuclei. It explains phenomena like radioactive decay, nuclear fission, and fusion. This field helps us understand the reactions that power both stars in the universe and the atomic bombs.

By studying nuclear physics, scientists can predict how elements behave over time, calculate energy release from nuclear reactions, and even develop medical treatments using radioactive materials. The study of half-life is a fundamental aspect of nuclear physics. It allows scientists to measure how fast a radioactive substance decays, which is pivotal in various fields from archaeology to power generation.
  • Scope: Studies atomic nuclei and interactions.
  • Applications: Power generation, medical treatments, understanding cosmic phenomena.
  • Key Concepts: Radioactive decay, fission, fusion, half-life calculations.
  • Impact: Vital to developing new technologies and advancing scientific knowledge.

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Most popular questions from this chapter

1 Fusion process, line combining two deutrons to form a He nucleus are impossible at ordinary temperatures and pressure. This reasons for this can be traced to the fact [NCERT Exemplar] (a) nuclear forces have short range (b) nuclei are positively charge (c) the original nuclei must be completely ionized before fussion can take place (d) the original nuclei must first break up before combining with each other.

The penetrating powers of \(\alpha, \beta\) and \(\gamma\) radiations, in decreasing order are (a) \(v, \alpha, \beta\) (b) \(\gamma, \beta, \alpha\) (c) \(\alpha, \beta, \gamma\) (d) \(\beta, \gamma, \alpha\)

The normal activity of living carbon containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive \({ }_{6}^{14} \mathrm{C}\) present with the stable carbon isotope \({ }_{6}^{12} \mathrm{C}\). when the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life \((5730 \mathrm{yr})\) of \({ }_{6}^{14} \mathrm{C}\) and the measured activity, the age of the specimen can be approximately estimated. This is the principle of \({ }_{6}^{14} \mathrm{C}\) dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilization. (a) \(5224 \mathrm{yr}\) (b) \(4224 \mathrm{yr}\) (c) \(8264 \mathrm{yr}\) (d) \(6268 \mathrm{yr}\)

An \(\alpha\)-particle of mass \(6.65 \times 10^{-27} \mathrm{~kg}\) travels at right angles to a magnetic field of \(0.2 \mathrm{~T}\) with a speed of \(6 \times 10^{5} \mathrm{~ms}^{-1}\). The acceleration of \(\alpha\)-particle will be (a) \(9.77 \times 10^{11} \mathrm{~ms}^{-2}\) (b) \(8.55 \times 10^{11} \mathrm{~ms}^{-2}\) (c) \(5.77 \times 10^{12} \mathrm{~ms}^{-2}\) (d) \(7.55 \times 10^{12} \mathrm{~ms}^{-2}\)

Light of wavelength \(488 \mathrm{~nm}\) is produced by an argon laser, which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is \(0.38 \mathrm{~V}\). Find the work function of the material from which the emitter is made. (a) \(2.2 \mathrm{eV}\) (b) \(3.7 \mathrm{eV}\) (c) \(1.6 \mathrm{eV}\) (d) \(4.2 \mathrm{eV}\)
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