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Chapter 29: Problem 66

Approximately what is the total energy of the neutrino emitted when ${ }_{11}^{22} \mathrm{Na}$ decays by electron capture?

Short Answer

Expert verified
Answer: The approximate total energy of the emitted neutrino is 2.85 MeV.

Step by step solution

01

Identify the daughter nucleus

After electron capture, a proton in the parent nucleus transforms into a neutron, thus decreasing the atomic number (Z) by 1 while keeping the mass number (A) constant. So, the daughter nucleus will have an atomic number of 10 and a mass number of 22, which is \({}_{10}^{22}\mathrm{Ne}\).
02

Calculate the mass difference between parent and daughter nuclei

We can find the mass difference by using the atomic mass values for the parent (\({m}_{\mathrm{Na}}\)) and daughter (\({m}_{\mathrm{Ne}}\)) nuclei. Using a reference database, we find the atomic mass values as follows: \({m}_{\mathrm{Na}} = 21.9944364\) amu \({m}_{\mathrm{Ne}} = 21.9913849\) amu The mass difference (\(螖m\)) is obtained by subtracting the daughter nucleus mass from the parent nucleus mass: \(螖m = {m}_{\mathrm{Na}} - {m}_{\mathrm{Ne}} = 21.9944364 - 21.9913849 = 0.0030515\) amu
03

Convert the mass difference to energy

Now, we'll use the energy-mass equivalence principle (E=mc虏) to convert the mass difference into energy. First, convert the mass difference from atomic mass units (amu) to kilograms (kg), using the conversion factor 1 amu = 1.66054 脳 10^{-27} kg: \(螖m_\mathrm{kg} = 0.0030515 \times 1.66054 脳 10^{-27} = 5.06895 脳 10^{-30}\) kg Next, apply the energy-mass equivalence principle with the speed of light (c) value as 3 脳 10^8 m/s: \(E = 螖m_\mathrm{kg} \times c^2 = 5.06895 脳 10^{-30} \times (3 脳 10^8)^2 = 4.5620 脳 10^{-13}\) Joules
04

Express the energy in a suitable unit and approximate the value

Nuclear reactions usually deal with energy in the unit of electronvolts (eV). So, we'll convert the calculated energy into MeV (Mega-electronvolts), using the conversion factor 1 Joule = 6.242 脳 10^{12} eV: \(E_\mathrm{MeV} = 4.5620 脳 10^{-13} \times 6.242 脳 10^{12} \times 10^{-6} = 2.847\) MeV Approximately, the total energy of the emitted neutrino in the electron capture decay of \({}_{11}^{22}\mathrm{Na}\) is 2.85 MeV.

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Most popular questions from this chapter

Once Rutherford and Geiger determined the charge-to mass ratio of the \(\alpha\) particle (Problem 71 ), they performed another experiment to determine its charge. An \(\alpha\) source was placed in an evacuated chamber with a fluorescent screen. Through a glass window in the chamber, they could see a flash on the screen every time an \(\alpha\) particle hit it. They used a magnetic field to deflect \(\beta\) particles away from the screen so they were sure that every flash represented an alpha particle. (a) Why is the deflection of a \(\beta\) particle in a magnetic field much larger than the deflection of an \(\alpha\) particle moving at the same speed? (b) By counting the flashes, they could determine the number of \(\alpha\) s per second striking the screen (R). Then they replaced the screen with a metal plate connected to an electroscope and measured the charge \(Q\) accumulated in a time \(\Delta t .\) What is the \(\alpha\) -particle charge in terms of \(R, Q,\) and \(\Delta t ?\)
Which of these unidentified nuclides are isotopes of each other? $$ { }_{71}^{175}(?) $$ $$ \frac{71}{32}(?) $$ $$ \frac{175}{74}(?) $$ $$ { }_{71}^{167}(?) $$ $$ \frac{71}{30}(?) $$ $$ { }_{74}^{180}(?) $$
An \(\alpha\) particle produced in radioactive \(\alpha\) decay has a kinetic energy of typically about 6 MeV. When an \(\alpha\) particle passes through matter (e.g., biological tissue), it makes ionizing collisions with molecules, giving up some of its kinetic energy to supply the binding energy of the electron that is removed. If a typical ionization energy for a molecule in the body is around \(20 \mathrm{eV}\) roughly how many molecules can the alpha particle ionize before coming to rest?
Show that the spontaneous \(\alpha\) decay of \(^{19} \mathrm{O}\) is not possible.
Radium- 226 decays as ${ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He} .\( If the \){ }_{88}^{226}$ Ra nucleus is at rest before the decay and the \({ }_{86}^{222} \mathrm{Rn}\) nucleus is in its ground state, estimate the kinetic energy of the \(\alpha\) particle. (Assume that the \({ }_{86}^{222} \mathrm{Rn}\) nucleus takes away an insignificant fraction of the kinetic energy.)
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