/*! 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 27 The isotope potassium-40 is a \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The isotope potassium-40 is a \(\beta\) -emitter. Write out the parentdaughter equation for the decay.

Short Answer

Expert verified
The parent-daughter equation is: \( ^{40}_{19}\text{K} \rightarrow ^{40}_{20}\text{Ca} + \beta^- + \overline{v}_e \).

Step by step solution

01

Identify the Parent Isotope

The parent isotope in this decay process is Potassium-40 (\(^ {40} \text{K}\)). Potassium-40 is a known \(\beta^-\)-emitter.
02

Determine the Type of Decay

In \(\beta^-\) decay, a neutron is converted into a proton, causing an electron (\(\beta^-\) particle) to be emitted. This results in the increase of the atomic number by 1 while the mass number remains the same.
03

Identify the Daughter Isotope

Upon \(\beta^-\) decay, Potassium-40 (\(^ {40} \text{K}\)) will convert to Calcium-40 (\(^ {40} \text{Ca}\)), since the atomic number increases from 19 (Potassium) to 20 (Calcium). The mass number remains at 40.
04

Write the Decay Equation

The parent-daughter equation for the decay can be written as follows: \[^{40}_{19}\text{K} \rightarrow \, ^{40}_{20}\text{Ca} + \beta^- + \overline{v}_e\,\]where \(^{40}_{19}\text{K}\) is the parent Potassium-40, \(^{40}_{20}\text{Ca}\) is the daughter Calcium-40, \(\beta^-\) is the beta particle, and \(\overline{v}_e\) is the antineutrino.

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.

Isotopes
Isotopes are fascinating variants of elements. They have the same number of protons but a different number of neutrons. This means isotopes of an element have the same atomic number but different mass numbers. For example, let's consider the element Potassium. It has several isotopes, including Potassium-39, Potassium-40, and Potassium-41. All these isotopes share 19 protons since Potassium has an atomic number of 19.
However, the number of neutrons differs:
  • Potassium-39 has 20 neutrons.
  • Potassium-40 has 21 neutrons.
  • Potassium-41 has 22 neutrons.
Despite these differences in neutron number, isotopes exhibit similar chemical behavior because their electron count remains unchanged due to the constant number of protons.
Nuclear Decay Equations
Nuclear decay equations are crucial for understanding how unstable isotopes transform into more stable ones. One key concept is that during decay, the sum of atomic masses and numbers must remain balanced. This adherence to conservation laws helps in predicting the products of decay processes.
There are several types of radioactive decay, such as alpha decay, beta decay, and gamma decay. For instance, in beta decay:
  • A neutron in the nucleus decays into a proton.
  • An electron (beta particle) and an antineutrino are emitted.
In nuclear decay equations, you express this transformation process. A parent isotope like Potassium-40 decays into a daughter isotope, Calcium-40, involving a change in the atomic number but not in the mass number. On paper, this is shown through equations that neatly tally both mass numbers and atomic numbers before and after the decay.
Potassium-40 Decay
Potassium-40 decay is a classic example of beta decay where a neutron in the nucleus is transformed into a proton. This type of decay is denoted as \(\beta^-\) decay. During this process, Potassium-40 (^ {40} \text{K}) undergoes a transition to form Calcium-40 (^ {40} \text{Ca}). In this change:
  • Potassium's atomic number increases from 19 to 20, making it Calcium.
  • The mass number remains unchanged at 40.
The equation representing this decay is\[^{40}_{19}\text{K} \rightarrow \, ^{40}_{20}\text{Ca} + \beta^- + \overline{v}_e\,\]where \(\beta^-\) is the beta particle, and \(\overline{v}_e\) is the antineutrino emitted during the process. Understanding this decay is key to grasping more advanced concepts in nuclear physics.
Physics Education
Learning physics, whether in a classroom or through self-study, involves exploring how fundamental concepts like energy, force, and matter behave and interact. Radioactivity, and specifically beta decay, provides an excellent context to learn these ideas in action. It's about understanding not just what happens during a decay process, but why it happens.
When students dive into topics such as isotopes and nuclear decay, they embrace complex but fascinating world forms beyond standard chemistry.
Through clear explanations and engaging exercises, students enhance their critical thinking and problem-solving skills. Using detailed animations and practice problems can further illustrate abstract concepts such as Potassium-40 decay to make physics more engaging and relatable for learners of all levels.

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

A nucleus \({ }_{n}^{M} \mathrm{P}\), the parent nucleus, decays to a daughter nucleus D by positron decay: $$ { }_{n}^{n} \rightarrow D+{ }_{+1}^{0} e+{ }_{0}^{0} v $$ where \(v\) is a neutrino, a particle that has nearly zero mass and zero charge. ( \(a\) ) What are the subscript and superscript for D? (b) Prove that the mass loss in the reaction is \(M_{p}-M_{d}-2 m_{e}\), where \(M_{p}\) and \(M_{d}\) are the atomic masses of the parent and daughter. (a) To balance the subscripts and superscripts, we must have \({ }_{n-1}^{M} \mathrm{D}\). (b) The table of masses for the nuclei involved is Subtraction gives the mass loss: $$ \left(M_{p}-n m_{e}\right)-\left(M_{d}-n m_{e}+2 m e\right)=M_{p}-M_{d}-2 m_{e} $$ Notice how important it is to keep track of the electron masses in this and the previous problem.

An experiment is done to determine the half-life of a radioactive substance that emits one beta particle for each decay process. Measurements show that an average of \(8.4\) beta particles are emitted each second by \(2.5 \mathrm{mg}\) of the substance. The atomic mass of the substance is \(230 .\) Find the half- life of the substance.

By natural radioactivity \({ }^{238} \mathrm{U}\) emits an \(\alpha\) -particle. The heavy residual nucleus is called \(\mathrm{UX}_{1} . \mathrm{UX}_{1}\) in turn emits a beta particle. The resultant nucleus is called \(\mathrm{UX}_{2} .\) Determine the atomic number and mass number for \((a) \mathrm{UX}_{1}\) and \((b) \mathrm{UX}_{2}\).

How many protons, neutrons, and electrons are there in \((a){ }^{3} \mathrm{He}\), (b) \(^{12} \mathrm{C}\), and \((c)^{206} \mathrm{~Pb}\) ? (a) The atomic number of He is 2; therefore, the nucleus must contain 2 protons. Since the mass number of this isotope is 3 , the sum of the protons and neutrons in the nucleus must equal 3; therefore, there is 1 neutron. The number of electrons in the atom is the same as the atomic number, 2 . (b) The atomic number of carbon is 6 ; hence, the nucleus must contain 6 protons. The number of neutrons in the nucleus is equal to \(12-6=6\). The number of electrons is the same as the atomic number, 6 . (c) The atomic number of lead is 82 ; hence, there are 82 protons in the nucleus and 82 electrons in the atom. The number of neutrons is \(206-82=124\).

Technetium-99 (æ?r) has an excited state that decays by emission of a gamma ray. The half-life of the excited state is 360 min. What is the activity, in curies, of \(1.00 \mathrm{mg}\) of this excited isotope? Because we have the half-life \(\left(t_{1 / 2}\right)\) we can determine the decay constant since \(\lambda t_{1 / 2}=0.693 .\) The activity of a sample is \(\lambda N .\) In this case, $$ \lambda=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{21600 \mathrm{~s}}=3.21 \times 10^{-5} \mathrm{~s}^{-1} $$ We also know that \(99.0 \mathrm{~kg}\) of Tc contains \(6.02 \times 10^{26}\) atoms. A mass \(m\) will therefore contain \([m /(99.0 \mathrm{~kg})]\left(6.02 \times 10^{26}\right)\) atoms. In our case, \(m=1.00 \times 10^{-6} \mathrm{~kg}\), and so $$ \begin{aligned} \text { Activity } &=\lambda N=\left(3.21 \times 10^{-5} \mathrm{~s}^{-1}\right)\left(\frac{1.00 \times 10^{-6} \mathrm{~kg}}{99.0 \mathrm{~kg}}\right)\left(6.02 \times 10^{26}\right) \\ &=1.95 \times 10^{14} \mathrm{~s}^{-1}=1.95 \times 10^{14} \mathrm{~Bq} \end{aligned} $$

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.