In each of the following radioactive decay processes, supply the missing particle. a. \({ }^{60} \mathrm{Co} \rightarrow{ }^{60} \mathrm{Ni}+\) ? b. \({ }^{97} \mathrm{Tc}+? \rightarrow{ }^{97} \mathrm{Mo}\) c. \({ }^{99} \mathrm{Tc} \rightarrow{ }^{99} \mathrm{Ru}+\) ? d. \({ }^{239} \mathrm{Pu} \rightarrow{ }^{235} \mathrm{U}+\) ?

Short Answer

Expert verified
a. \( { }^{60} \mathrm{Co} \rightarrow{ }^{60} \mathrm{Ni}+ \beta^- + \bar{\nu_e}\) b. \( { }^{97} \mathrm{Tc}+ \beta^- + \bar{\nu_e} \rightarrow{ }^{97} \mathrm{Mo}\) c. \( { }^{99} \mathrm{Tc} \rightarrow{ }^{99} \mathrm{Ru}+ \beta^- + \bar{\nu_e}\) d. \( { }^{239} \mathrm{Pu} \rightarrow{ }^{235} \mathrm{U}+ { }^4 \mathrm{He}\)

Step by step solution

01

Identify the type of decay

This process shows the conversion of Cobalt-60 to Nickel-60, which indicates a beta-minus decay (a neutron converting into a proton). The missing particle will be an electron (also called a beta particle) and an antineutrino.
02

Write the full decay process

With the identified missing particles, we can write the full decay process as: \[{ }^{60} \mathrm{Co} \rightarrow{ }^{60} \mathrm{Ni}+ \beta^- + \bar{\nu_e}\] #b. Technetium-97 decay#
03

Identify the type of decay

This process shows the conversion of Technetium-97 to Molybdenum-97. The missing particle will be a beta particle and an antineutrino since this decay process is for a shorter half-life period.
04

Write the full decay process

With the identified missing particles, we can write the full decay process as: \[{ }^{97} \mathrm{Tc}+ \beta^- + \bar{\nu_e} \rightarrow{ }^{97} \mathrm{Mo}\] #c. Technetium-99 decay#
05

Identify the type of decay

In this process, Technetium-99 decays into Ruthenium-99, indicating a beta-minus decay (a neutron converting into a proton). The missing particle will be an electron (beta particle) and an antineutrino.
06

Write the full decay process

With the identified missing particles, we can write the full decay process as: \[{ }^{99} \mathrm{Tc} \rightarrow{ }^{99} \mathrm{Ru}+ \beta^- + \bar{\nu_e}\] #d. Plutonium-239 decay#
07

Identify the type of decay

In this decay process, the conversion of Plutonium-239 to Uranium-235 indicates alpha decay (emission of an alpha particle, which consists of 2 protons and 2 neutrons). There is no additional particle involved in this process.
08

Write the full decay process

With the identified missing particle (alpha particle), we can write the full decay process as: \[{ }^{239} \mathrm{Pu} \rightarrow{ }^{235} \mathrm{U}+ { }^4 \mathrm{He}\]

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!

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

Estimate the temperature needed to achieve the fusion of deuterium to make an \(\alpha\) particle. The energy required can be estimated from Coulomb's law [use the form \(E=9.0 \times 10^{9}\) \(\left(Q_{1} Q_{2} / r\right)\), using \(Q=1.6 \times 10^{-19} \mathrm{C}\) for a proton, and \(r=2 \times\) \(10^{-15} \mathrm{~m}\) for the helium nucleus; the unit for the proportionality constant in Coloumb's law is \(\left.\mathrm{J} \cdot \mathrm{m} / \mathrm{C}^{2}\right]\).

One type of commercial smoke detector contains a minute amount of radioactive americium- \(241\left({ }^{241} \mathrm{Am}\right)\), which decays by \(\alpha\) -particle production. The \(\alpha\) particles ionize molecules in the air, allowing it to conduct an electric current. When smoke particles enter, the conductivity of the air is changed and the alarm buzzes. a. Write the equation for the decay of \({ }_{95}^{241} \mathrm{Am}\) by \(\alpha\) -particle production. b. The complete decay of \({ }^{241}\) Am involves successively \(\alpha, \alpha, \beta\), \(\alpha, \alpha, \beta, \alpha, \alpha, \alpha, \beta, \alpha\), and \(\beta\) production. What is the final stable nucleus produced in this decay series? c. Identify the 11 intermediate nuclides.

A rock contains \(0.688 \mathrm{mg}^{206} \mathrm{~Pb}\) for every \(1.000 \mathrm{mg}{ }^{238} \mathrm{U}\) present. Assuming that no lead was originally present, that all the \({ }^{206} \mathrm{~Pb}\) formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between \({ }^{238} \mathrm{U}\) and \({ }^{206} \mathrm{~Pb}\) is negligible, calculate the age of the rock. (For \({ }^{238} \mathrm{U}\), \(t_{1 / 2}=4.5 \times 10^{9}\) years. \()\)

The only stable isotope of fluorine is fluorine- \(19 .\) Predict possible modes of decay for fluorine- 21 , fluorine- 18 , and fluorine- 17 .

Many elements have been synthesized by bombarding relatively heavy atoms with high-energy particles in particle accelerators. Complete the following nuclear reactions, which have been used to synthesize elements. a. _______ \(+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{97}^{243} \mathrm{Bk}+{ }_{0}^{1} \mathrm{n}\) b. b. \({ }_{92}^{238} \mathrm{U}+{ }_{6}^{12} \mathrm{C} \rightarrow\) ______ \(+6{ }_{0}^{1} \mathrm{n}\) c. C. \({ }_{98}^{249} \mathrm{Cf}+\) _____ \(\longrightarrow{ }_{105}^{260} \mathrm{Db}+4{ }_{0}^{1} \mathrm{n}\) d. \({ }_{98}^{249} \mathrm{Cf}+{ }_{5}^{10} \mathrm{~B} \rightarrow{ }_{103}^{257} \mathrm{Lr}+\) ______.

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