/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 The half-life of the hydrogen is... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The half-life of the hydrogen isotope tritium is about 12 years. After a certain amount of time a fraction \(31 / 32\) of the atoms in the original sample has decayed. The time is most nearly equal to (A) 12 years (B) 24 years (C) 36 years (D) 48 years (E) 60 years

Short Answer

Expert verified
Answer: 60 years

Step by step solution

01

Write down the formula for exponential decay

The formula for exponential decay is: \(N(t) = N_0 e^{-\lambda t}\), where: \(N(t)\): number of the substance left after time \(t\), \(N_0\): initial amount of the substance, \(\lambda\): decay constant, \(t\): time elapsed.
02

Relate half-life to the decay constant \(\lambda\)

To relate the half-life \(T_{1/2}\) to the decay constant \(\lambda\), use the formula: \(T_{1/2} = \frac{\ln{2}}{\lambda}\).
03

Find the decay constant \(\lambda\) for tritium

We are given that the half-life of tritium is 12 years. Using the formula from Step 2, we can find the decay constant \(\lambda\): \(\lambda = \frac{\ln{2}}{T_{1/2}} = \frac{\ln{2}}{12}\).
04

Write down the fraction of tritium decayed and set up an equation

The fraction decayed is given to be \(31/32\). Therefore, the fraction remaining (not decayed) is \(1 - \frac{31}{32} = \frac{1}{32}\). Now we can set up an equation for the remaining fraction of tritium: \(\frac{N(t)}{N_0} = \frac{1}{32} = e^{-\lambda t}\).
05

Solve for time \(t\)

Substitute the value of \(\lambda\) found in Step 3 into the equation from Step 4: \(\frac{1}{32} = e^{-\frac{\ln{2}}{12} t}\). To solve for \(t\), we can take the natural logarithm of both sides: \(\ln{\frac{1}{32}} = -\frac{\ln{2}}{12}t\). Now, we can solve for \(t\): \(t = \frac{12 \ln{(\frac{1}{32})}}{-\ln{2}} \approx 60\). So, the time elapsed is most nearly equal to 60 years, which is option (E).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Half-Life Calculation
Understanding the concept of half-life is crucial in fields like nuclear physics, chemistry, and environmental science. A half-life is the time required for a quantity to reduce to half its initial value. In radioactive decay, it's the time it takes for half of the radioactive atoms in a sample to disintegrate.

To calculate the half-life when given a decay constant \(\lambda\), we use the formula: \[ T_{1/2} = \frac{\ln{2}}{\lambda} \]. Knowing the half-life helps us understand the longevity and decay rate of radioactive substances.

Step-by-Step Approach to Half-Life Problems

Given the decay constant or the half-life, one can determine the remaining amount of a substance after any period using the right equations. It's a basic yet powerful tool in solving exponential decay problems.
Tritium Decay
Tritium, or hydrogen-3, is a radioactive isotope with notable use in self-illuminating devices and as a tracer in environmental studies. Its decay is a great example to explore the principles of radioactivity. Tritium decays via beta emission, transforming into helium-3 with a relatively short half-life of about 12 years.

Due to its decay properties, tritium is an excellent candidate for half-life studies in physics and chemistry education.

Decay Representation

The decay of tritium can be represented in equations that express the initial and remaining quantities of the isotope over time, helping us predict the amount present after a specific period.
Radioactive Decay Equations
Radioactive decay is mathematically represented by equations that describe how unstable atoms lose energy by emitting radiation. The general formula for exponential decay of a radioactive substance is: \[ N(t) = N_0 e^{-\lambda t} \], where \(N(t)\) is the number of atoms remaining after time \(t\), \(N_0\) is the initial number of atoms, \(\lambda\) is the decay constant, and \(t\) is the elapsed time.

This equation is vital for understanding how radioactive substances change over time, and is a fundamental piece of knowledge for students studying nuclear physics.

Application in Real-World Scenarios

By applying these equations, scientists and engineers can manage nuclear materials, date archaeological finds, and ensure safety in medical treatments involving radioactive isotopes.
AP Physics Practice
Engaging with problems like the tritium decay example is perfect for AP Physics practice. These problems encompass a variety of concepts including exponential functions, logarithms, and of course, physics. They encourage critical thinking and a deep understanding of physical principles.

The practice of working through such exercises prepares students for the problem-solving nature of AP examinations.

Essential Skills

Through these exercises, students develop skills in equation manipulation, exponential decay understanding, and application of logarithmic functions, all while reinforcing their knowledge of the subject.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An atomic nucleus is induced to break into two pieces, during which process energy is released. One can say with certainty that (A) the original atomic mass was greater than that of iron. (B) the fragments will each have an atomic mass less than that of iron. (C) the masses of the fragments will add up to be less than the mass of the original nucleus. (D) (A) and (C) (E) None of the above

Which of the following statements about the photon is true? i. The photon has a wavelength; ii. The photon has a mass; iii. The photon can undergo interference; iv. The photon is charged; \(\mathrm{v}\). The photon has a momentum. (A) i only (B) i, ii only (C) i, iii, v only (D) i, ii, iii, iv only (E) iii, v only

In the Bohr model of the atom, which of the following statements are false? (More than one response permitted.) (A) The energy levels are spaced apart by \(h v\). (B) The angular momentum of the electron is quantized. (C) The least amount of energy that can be absorbed by the atom is between the \(n=2\) state and the \(n=3\) state. (D) The energy of a photon emitted in the Balmer series is proportional to \(\left(\frac{1}{4}-\frac{1}{n^2}\right)\) where \(n\) is the principal quantum number of the initial energy level. (E) The energy levels characterized by the quantum number \(n\) are stable.

A photon of energy \(1.75 \mathrm{eV}\) has a wavelength of most nearly (A) \(710 \mathrm{~nm}\) (B) \(700 \mathrm{~nm}\) (C) \(650 \mathrm{~nm}\) (D) \(600 \mathrm{~nm}\) (E) \(550 \mathrm{~nm}\)

The nucleus of an ordinary hydrogen atom consists of (A) a neutron. (B) a proton. (C) a proton and a neutron. (D) a proton and two neutrons. (E) two protons and two neutrons.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.