The radionuclideIr196decays by emitting an electron. (a) Into which square in Fig. 42-6 is it transformed? (b) Do further decays then occur?

Short Answer

Expert verified

a) It is transformed into beige square in Fig.42-6.

b) Further decays do occur.

Step by step solution

01

Write given data

In Fig.42-6, the radioactive decay of nuclide Ir196by emitting an electron is given.

02

Determine the concept of decay  

In beta decay, a nuclide decays while emitting an electron with a neutrino particle. This implies that the proton number of the nuclide is added by 1 more proton.

03

a) Calculate the square of the transformed radionuclide

In the figure 42-6, observe that the radionuclideIr196 is present in the beige square that resembles it radioactive nature with the given half-life of 52s.

04

b) Determine whether it continues decaying

By emitting an electron, the radionuclideIr196 undergoes a beta decay process. That can be given as follows:

Ir196Pt197+e-1+v

The product of the decay process is a radionuclide Pt197, which also lies in the beige region with a half-life of 18.3h.

Hence, even after the decay process, the decay still occurs.

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

Suppose the alpha particle in a Rutherford scattering experiment is replaced with a proton of the same initial kinetic energy and also headed directly toward the nucleus of the gold atom. (a) Will the distance from the center of the nucleus at which the proton stops be greater than, less than, or the same as that of the alpha particle? (b) If, instead, we switch the target to a nucleus with a larger value of Z,is the stopping distance of the alpha particle greater than, less than or the same as with the gold target?

How much energy is released when a 238nucleus decays by emitting (a) an alpha particle and (b) a sequence of neutron, proton, neutron, and proton? (c) Convince yourself both by reasoned argument and by direct calculation that the difference between these two numbers is just the total binding energy of the alpha particle. (d) Find that binding energy. Some needed atomic and particle masses are

U238238.05079uT234h234.04363uU237237.04873uH4e4.00260uU236236.04891uH11.00783uU235235.04544un1.00866u

The radioactive nuclideTc99 can be injected into a patient’s bloodstream in order to monitor the blood flow, measure the blood volume, or find a tumor, among other goals. The nuclide is produced in a hospital by a “cow” containingM99o, a radioactive nuclide that decays toT99cwith a half-life of 67h. Once a day, the cow is “milked” for itsT99c, which is produced in an excited state by theM99o; theT99cde-excites to its lowest energy state by emitting a gamma-ray photon, which is recorded by detectors placed around the patient. The de-excitation has a half-life of 6.0h. (a) By what process does99Modecay to99Tc? (b) If a patient is injected with a8.2×107Bqsample of99Tc, how many gamma-ray photons are initially produced within the patient each second? (c) If the emission rate of gamma-ray photons from a small tumor that has collected99Tc is 38 per second at a certain time, how many excited states99Tcare located in the tumor at that time?

A radiation detector records 8700 counts in 1.00 min. Assuming that the detector records all decays, what is the activity of the radiation source in (a) Becquerel and (b) curies?

Some radionuclides decay by capturing one of their own atomic electrons, a K-shell electron, say. An example is

49V+e-49Ti+v,T1/2=331d

Show that the disintegration energy Qfor this process is given by

Q=(mv-mTi)c2-EK

where,mvandmTiare the atomic masses of49Vand49Ti, respectively, andEKis the binding energy of the vanadium K-shell electron. (Hint:Putandas the corresponding nuclear masses and then add in enough electrons to use the atomic masses.)

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.

Sign-up for free