Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Silicon-31 has a half-life of approximately 2.5 hours. If we begin with a sample containing \(1000 \mathrm{mg}\) of Si-31, what is the approximate amount remaining after 10 hours?

Short Answer

Expert verified
After 10 hours, there will be approximately 62.5 mg of Si-31 remaining from the initial 1000 mg sample.

Step by step solution

01

Listing down given information

We have the following information to solve this problem: 1. Half-life of Si-31 (T_half) = 2.5 hours 2. Initial amount of Si-31 (N_initial) = 1000 mg 3. Time elapsed (t) = 10 hours
02

Understand the half-life formula

To determine the remaining amount of Si-31, we will use the half-life formula: \[N_t = N_0 * (1/2)^\frac{t}{T_{half}}\] Where: - \(N_t\) is the amount remaining after time 't' - \(N_0\) is the initial amount - \(t\) is the time elapsed - \(T_{half}\) is the half-life of the substance
03

Substitute the given information into the formula

Now, we will plug the values into the formula: \[N_t = 1000 * (1/2)^\frac{10}{2.5}\]
04

Solve for the remaining amount of Si-31

Calculate the result: \(N_t = 1000 * (1/2)^{4}\) \(N_t = 1000 * (1/16)\) \(N_t = 62.5 \mathrm{mg}\)
05

Interpret the result

After 10 hours, there will be approximately 62.5 mg of Si-31 remaining from the initial 1000 mg sample.

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 Vaia!

Key Concepts

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

Silicon-31 Decay
Silicon-31 (\( \text{Si-31} \)) is a radioactive isotope of silicon. It undergoes decay, which means it gradually loses its radioactivity over time. This process occurs at a predictable rate, known as its half-life. For Si-31, the half-life is about 2.5 hours. This specific timeframe indicates the period required for half of any given sample of Si-31 to decay.
  • If you start with 1000 mg, after 2.5 hours, only 500 mg would be remaining.
  • After another 2.5 hours (total 5 hours), this amount would half again to 250 mg.
  • This continuous process of halving continues over successive half-lives.
Silicon-31's steady decay pattern helps scientists predict how long it takes for a sample of this isotope to reach a desired amount. This predictable nature of decay is essential in fields like radiometric dating and nuclear medicine, where specific isotopes are used for diagnostic or treatment purposes.
Radioactive Decay Formula
The radioactive decay of a substance can be mathematically understood using the half-life formula. This formula is critical for calculating the amount of a radioactive material that remains after a certain period:\[N_t = N_0 \times \left( \frac{1}{2} \right)^\frac{t}{T_{half}}\]Here's what each symbol stands for:
  • \(N_t\): the amount of substance remaining after time \(t\).
  • \(N_0\): the starting amount of the substance.
  • \(t\): the time elapsed.
  • \(T_{half}\): the half-life of the substance.
This formula captures the essence of exponential decay. Every time one half-life passes, the amount of the substance is halved. It provides a straightforward way to compute how much of a radioactive isotope remains, simplifying complex radioactive decay processes into easy-to-understand math. Understanding this formula is key to mastering half-life calculations and applying them effectively in scientific problems.
Amount Remaining Calculation
Determining how much of a radioactive substance remains after a specified period involves plugging values into the half-life decay formula. Let's go through the process using Silicon-31:
  • The initial mass (\(N_0\)) is 1000 mg.
  • The half-life (\(T_{half}\)) is 2.5 hours.
  • The elapsed time (\(t\)) is 10 hours.
  • Use the formula: \[ N_t = N_0 \times \left( \frac{1}{2} \right)^\frac{t}{T_{half}} \]
Substitute the known values:\[ N_t = 1000 \times \left( \frac{1}{2} \right)^\frac{10}{2.5} \]Now solve the exponent:
  • The exponent is \( \frac{10}{2.5} = 4 \).
  • This means we halve the original mass four times in total.
Calculation:
  • First halve: 1000 mg \( \times \frac{1}{2} = 500 \) mg
  • Second halve: 500 mg \( \times \frac{1}{2} = 250 \) mg
  • Third halve: 250 mg \( \times \frac{1}{2} = 125 \) mg
  • Fourth halve: 125 mg \( \times \frac{1}{2} = 62.5 \) mg
After 10 hours, you will find about 62.5 mg of Si-31 remains. Understanding this step-by-step breakdown helps in grasping how exponential decay works and makes calculating the remaining amount straightforward and logical.

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

During the research that led to production of the two atomic bombs used against Japan in World War II, different mechanisms for obtaining a supercritical mass of fissionable material were investigated. In one type of bomb, what is essentially a gun was used to shoot one piece of fissionable material into a cavity containing another piece of fissionable material. In the second type of bomb, the fissionable material was surrounded with a high explosive that, when detonated, compressed the fissionable material into a smaller volume. Discuss what is meant by critical mass, and explain why the ability to achieve a critical mass is essential to sustaining a nuclear reaction.

Complete each of the following nuclear equations by supplying the missing particle. a. \(? \rightarrow_{84}^{210} \mathrm{Po}+_{2}^{4} \mathrm{He}\) b. \(40 \mathrm{K} \rightarrow_{18}^{40} \mathrm{Ar}+?\) c. \(\frac{137}{57} \mathrm{La}+_{-1} \mathrm{e} \rightarrow ?\)

Define the terms nuclear fission and nuclear fusion. Which process results in the production of a heavier nucleus? Which results in the production of smaller nuclei?

Aluminum exists in several isotopic forms, includ\(\operatorname{ing}_{13}^{27} \mathrm{Al}, \underset{13}{28} \mathrm{Al},\) and \(_{13}^{29} \mathrm{Al} .\) Indicate the number of protons and the number of neutrons in each of these isotopes.

Write a balanced nuclear equation for the decay of each of the following nuclides to produce a beta particle. a. \(\quad \frac{136}{53}\)I b. \(\quad \frac{133}{51} S b\) c. \(\quad \frac{117}{49} \mathrm{In}\)

See all solutions

Recommended explanations on Chemistry 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.

Sign-up for free