{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1 - positron silicon-28 mass number 28 28 atomic number 15...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
The symbol for phosphorus-28 = . The atomic weight of phosphorus is 31.0 . The atomic mass of phosphorus- 28 is about 28 . Therefore the nuclide is light compared to the stable isotopes of phosphorus . This means that the isotope has too few neutrons, or that its neutron/proton ratio is too low . It will decay to increase this ratio. This can be accomplished by either electron capture or positron emission, in which the mass number remains constant while the atomic number decreases. The nuclide with the same mass number but with an atomic number of 15 - 1 = 14 is silicon-28 = . Lighter atoms are more likely to decay by positron emission than by electron capture. In this case the decay is by positron emission. The reaction is: + Check: For both mass and charge to be conserved in a nuclear reaction: (1) The sum of the mass numbers for the products must equal the sum of the mass numbers for the reactants. (2) The sum of the atomic numbers for the products must equal the sum of the atomic numbers for the reactants. phosphorus-28
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: positron + silicon-28 + mass number: 28 + 28 atomic number: 15 1 + 14 Since radioactive decay is first order, the integrated rate law for a radioactive decay reaction is given by ln(N/N o ) = -kt where N o is the original number of radioactive nuclides, N is the number that remain at time t, and k is the rate constant for the reaction. The relation between the rate constant and the half-life for a first order reaction is kt 1/2 = ln 2 = 0.693 (1) Calculate the rate constant from the half-life given: half-life = t 1/2 = 27.7 days k = (ln 2 / t 1/2 ) = (0.693 / 27.7 days ) = 2.502E-2 d-1 (2) Calculate the time required for the activity to fall to 8.87 percent of its original value: The activity is directly proportional to the number or radioactive nuclides, N. Let N o be the original number of nuclides. N = 8.87E-2 x N o , where 8.87E-2 = 8.87 / 100 Then N/N o = 8.87E-2 Rearranging ln(N/N o ) = -kt t = -(1/k) x ln(N/N o ) = -(1 / 2.502E-2d-1 ) x ln ( 8.87E-2 ) = 96.8 d...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern