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

Element \(X\) is radioactive and decays via \(\alpha\) decay with a half-life of four days. If 12.5 percent of an original sample of element \(x\) remains after \(n\) days, what is the value of \(n ?\) (A) 4 (B) 8 (C) 12 (D) 16

Short Answer

Expert verified
n is 12 days, so the correct answer is (C).

Step by step solution

01

- Understand Half-life

Half-life is the time it takes for half of a sample of a radioactive element to decay. In this problem, the half-life of element X is given as 4 days.
02

- Determine remaining fraction

Since 12.5% (or 0.125) of the original sample remains, we need to determine how many half-lives (cycles of decay) result in this remaining fraction.
03

- Use the formula for decay

The remaining amount of a sample after n half-lives can be calculated using the formula: \[ \text{Remaining fraction} = \frac{1}{2^n} \] In this problem, \[ \frac{1}{2^n} = 0.125 \]
04

- Solve for n

To find n, solve the equation: \[ 0.125 = \frac{1}{2^n} \] We know that \[ 0.125 = \frac{1}{8} \] and since \[ 8 = 2^3 \], it follows that \( n = 3 \).
05

- Calculate total time in days

Each half-life cycle (n = 1) is 4 days. Since we have determined there are 3 half-lives, multiply these together to get the total time elapsed: \[ n \times \text{half-life} = 3 \times 4 = 12 \text{ days} \]
06

- Verify the answer

The value matches choice (C) in the problem statement. This confirms that after 12 days, 12.5% of the original sample remains.

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.

Half-life
Half-life is a key concept in understanding radioactive decay. It is the time required for half of the atoms in a radioactive sample to decay. For example, if you start with 100 atoms of element X and its half-life is 4 days, after 4 days, you would have 50 atoms left. This process continues, so after another 4 days (8 days total), you would have 25 atoms. Over time, the number of undecayed atoms decreases exponentially. Half-life helps in predicting how quickly a radioactive substance will diminish. When solving problems involving half-life, always remember that the time it takes for each successive halving remains constant.
Alpha Decay
Alpha decay is a type of radioactive decay where an unstable atom emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, exactly like a helium nucleus. This emission results in the atom losing 4 units of mass and 2 units of positive charge; it changes the parent atom to a different element. For instance, if element X undergoes alpha decay, it loses an alpha particle and transforms into a new element, often changing its chemical behavior. Understanding alpha decay is important to grasp how certain elements evolve over time and solve related problems.
Decay Formula
The decay formula is used to calculate the remaining amount of a radioactive substance over time. It's typically represented as:  \[ \text{Remaining Fraction} = \frac{1}{2^n} \] where 'n' is the number of half-life periods that have elapsed. This formula helps in determining how much of a substance remains after a certain number of half-lives. For example, if you start with 100 grams of element X and it has a half-life of 4 days, using the formula, after 8 days (2 half-lives), you would have: \[ \frac{1}{2^2} = \frac{1}{4}, \] so you’d have 25 grams left. By plugging in different values, you can predict how much of a substance remains at any given time.
Exponential Decay
Exponential decay describes how the quantity of a radioactive substance decreases over time in a predictable pattern. This type of decay occurs continuously and rapidly at first, then slows down. In mathematical terms, exponential decay can be explained using the formula: \[ N(t) = N_0 e^{-\text{λ}t}, \] where: \ * N(t) is the quantity remaining at time t, \ * N_0 is the initial quantity, \ * e is the base of the natural logarithm, \ * λ (lambda) is the decay constant, and \ * t is the time elapsed. While exponential decay may look daunting, the key takeaway is that it helps predict how quickly a sample diminishes over time. The decay constant λ can be related to the half-life with the formula: \[ \text{λ} = \frac{\text{ln}(2)}{\text{half-life}}. \] Understanding these relationships helps solve various problems involving radioactive decay predictably.

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

Which of the following correctly identifies the following process? $$_{31}^{67} \mathrm{Ga}+e^{-} \rightarrow_{30}^{67} \mathrm{Zn}$$ A. \(\beta^{-}\) decay B. \(\beta^{+}\) decay C. \(e^{-}\) capture D. \(\gamma\) decay

Consult your online resources for additional practice. If the work function of a metal is \(6.622 \times 10^{-20} \mathrm{J}\) and a ray of electromagnetic radiation with a frequency of \(1.0 \times 10^{14} \mathrm{Hz}\) is incident on the metal, what will be the speed of the electrons ejected from the metal? (Note: \(h=6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\) and \(\mathrm{m}_{\mathrm{e}^{-}}=9.1 \times\) \(\left.10^{-31} \mathrm{kg}\right)\) A. \(2.62 \times 10^{-6} \frac{\mathrm{m}}{\mathrm{s}}\) B. \(2.62 \times 10^{-6} \frac{\mathrm{m}}{\mathrm{s}}\) C. \(3.00 \times 10^{8} \frac{\mathrm{m}}{\mathrm{s}}\) D. \(3.00 \times 10^{8} \frac{\mathrm{m}}{\mathrm{s}}\)

What is the wavelength of a photon that causes an electron to be emitted from a metal with a kinetic energy of \(50 \mathrm{J} ?\) (Note: The work function of the metal is \(16 \mathrm{J}\), and \(h=6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\) ) A. \(1.0 \times 10^{-34} \mathrm{m}\) B. \(3.0 \times 10^{-27} \mathrm{m}\) C. \(3.0 \times 10^{-26} \mathrm{m}\) D. \(1.0 \times 10^{35} \mathrm{m}\)

A graph of an exponential decay process is created. The \(y\) -axis is the natural logarithm of the ratio of the number of intact nuclei at a given time to the number of intact nuclei at time \(t=0 .\) The \(x\) -axis is time. What does the slope of such a graph represent? A. \(\lambda\) B. \(-\lambda\) C. \(e^{-\lambda t}\) D. \(\frac{n}{n_{0}}\)

A nuclide undergoes two alpha decays, two positron decays, and two gamma decays. What is the difference between the atomic number of the parent nuclide and the atomic number of the daughter nuclide? A. 0 B. 2 C. 4 D. 6

See all solutions

Recommended explanations on English 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