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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 17 - Problem 90p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 17 - Problem 90p

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# Write a series of nuclear equations in wTiich Al-27 reacts with a neutron and the

ISBN: 9780321910295 34

## Solution for problem 90P Chapter 17

Introductory Chemistry | 5th Edition

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Problem 90P

Write a series of nuclear equations in which $$Al-27$$ reacts with a neutron and the product undergoes an alpha decay followed by a beta decay.

Equation Transcription:

Text Transcription:

Al-27

Step-by-Step Solution:
Step 1 of 3

#### Explanation:

So, you know that the first nuclear equation has aluminium-27 react with a neutron.

A neutron has no charge and an atomic mass equal to one, so you can write

27 13Al+10n→28 13Al

The resulting aluminium-28 isotope will then decay via alpha decay.

When isotopes undergo alpha decay, their nucleus emits an alpha particle, which is essentially the nucleus of a helium-4 atom.

Helium-4 has an atomic number equal to 2 and a mass number equal to 4. This means that you can write

28 13Al→AZX+4 2He

Since atomic number and mass number must be conserved in nuclear equations, you get that element X will have

A=28−4=24  and  Z=13−2=11

The element that has an atomic number equal to 11 is sodium, Na, which means that you are dealing with the sodium-24 isotope

28 13Al→24 11Na+4 2He

Finally, the sodium-24 isotope undergoes beta decay. In sodium-24's case, it's beta minus, β− decay.

During beta decay, a neutron from the nucleus of the radioactive isotope is converted into a proton. In addition to this, an electron and an electron antineutrino, ¯νe, are emitted by the nucleus.

So, if a neutron is converted into a proton, you can expect the atomic number to increase by 1, but the mass number to remain unchanged.

This means that you have

24 11Na→24 12Y+0-1e+¯νe

The element that has 12 protons in its nucleus is magnesium, which means that the final product of this decay chain is the magnesium-24 isotope.

24 11Na→24 12 Mg+0-1e+¯νe

Step 2 of 3

Step 3 of 3

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