/*! 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 39 Calculate the activity of \(1.0 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 29: Problem 39

Calculate the activity of \(1.0 \mathrm{g}\) of radium- 226 in Ci.

Short Answer

Expert verified
Answer: The activity of 1.0 g of radium-226 is approximately 0.9886 Ci.

Step by step solution

01

1. Determine the number of moles of radium-226 and convert it to atoms

First, we must determine the number of moles of radium-226 (Ra-226) in 1.0 g. To do this, we will use the molar mass of Ra-226, which is 226.0254 g/mol: Number of moles = mass / molar mass Number of moles of Ra-226 = 1.0 g / 226.0254 g/mol 鈮� 0.004424 mol Now we can convert the number of moles to atoms using Avogadro's number (6.022 x 10^23 atoms/mol): Number of atoms of Ra-226 = 0.004424 mol x 6.022 x 10^23 atoms/mol 鈮� 2.661 x 10^21 atoms
02

2. Calculate the decay constant for radium-226

To find the activity, we need the decay constant, which can be calculated using the half-life of radium-226. The half-life of radium-226 is 1600 years. We will first convert the half-life into seconds: Half-life in seconds = 1600 years x 365.25 days/year x 24 hours/day x 60 minutes/hour x 60 seconds/minute 鈮� 5.049 x 10^10 seconds Now, we can calculate the decay constant (位) using the formula \(\lambda = \frac{ln(2)}{T_{1/2}}\): 位 鈮� ln(2) / (5.049 x 10^10 seconds) 鈮� 1.373 脳 10^-11 s^-1
03

3. Determine the activity of radium-226 in becquerels (Bq)

Now we can calculate the activity in becquerels (Bq), which is equal to the number of decays per second. The activity (A) can be calculated using the formula A = 位N, where N is the number of atoms and 位 is the decay constant: A 鈮� (1.373 脳 10^-11 s^-1)(2.661 x 10^21 atoms) 鈮� 3.657 x 10^10 Bq
04

4. Convert the activity from Bq to Curies (Ci)

Finally, we need to convert the activity from becquerels (Bq) to Curies (Ci). The conversion factor is 1 Ci = 3.7 x 10^10 Bq, so we divide the activity in Bq by 3.7 x 10^10 Bq/Ci: Activity in Ci 鈮� 3.657 x 10^10 Bq / (3.7 x 10^10 Bq/Ci) 鈮� 0.9886 Ci So, the activity of 1.0 g of radium-226 is approximately 0.9886 Ci.

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 91影视!

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

Identify the daughter nuclide when \(_{19}^{40} \mathrm{K}\) decays via \(\beta^{-}\) decay.
Thorium-232 ( \(^{232}\) Th) decays via \(\alpha\) decay. Write out the reaction and identify the daughter nuclide.
The radioactive decay of \(^{238} \mathrm{U}\) produces \(\alpha\) particles with a kinetic energy of \(4.17 \mathrm{MeV} .\) (a) At what speed do these \(\alpha\) particles move? (b) Put yourself in the place of Rutherford and Geiger. You know that \(\alpha\) particles are positively charged (from the way they are deflected in a magnetic field). You want to measure the speed of the \(\alpha\) particles using a velocity selector. If your magnet produces a magnetic field of \(0.30 \mathrm{T},\) what strength electric field would allow the \(\alpha\) particles to pass through undeflected? (c) Now that you know the speed of the \(\alpha\) particles, you measure the radius of their trajectory in the same magnetic field (without the electric field) to determine their charge-to-mass ratio. Using the charge and mass of the \(\alpha\) particle, what would the radius be in a \(0.30-\mathrm{T}\) field? (d) Why can you determine only the charge-to-mass ratio \((q / m)\) by this experiment, but not the individual values of \(q\) and \(m ?\)
What is the average binding energy per nucleon for ${ }_{11}^{23} \mathrm{Na} ?$
An \(\alpha\) particle with a kinetic energy of \(1.0 \mathrm{MeV}\) is headed straight toward a gold nucleus. (a) Find the distance of closest approach between the centers of the \(\alpha\) particle and gold nucleus. (Assume the gold nucleus remains stationary. since its mass is much larger than that of the \(\alpha\) particle, this assumption is a fairly good approximation.) (b) Will the two get close enough to "touch"? (c) What is the minimum initial kinetic energy of an \(\alpha\) particle that will make contact with the gold nucleus?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Magnetism

Read Explanation

Dynamics

Read Explanation

Engineering Physics

Read Explanation

Radiation

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation
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.