Nuclear Chemistry

Just the basics….ByJ.M.Soltmann

What is Nuclear Chemistry

As its name implies, nuclear chemistry is the study of the nucleus and reactions between nuclei.

Remember that virtually all of the mass of an atom resides in the nucleus, as does all of the positive charge.

Nuclear energy is a much greater form of energy than bond energy.

As its name implies, nuclear chemistry is the study of the nucleus and reactions between nuclei.

Remember that virtually all of the mass of an atom resides in the nucleus, as does all of the positive charge.

Nuclear energy is a much greater form of energy than bond energy.

Radioactivity

While most nuclei are stable, many nuclei are unstable and spontaneously emit particles and electromagnetic radiation.

These nuclei are refered to as radionuclides.

While most nuclei are stable, many nuclei are unstable and spontaneously emit particles and electromagnetic radiation.

These nuclei are refered to as radionuclides.

Nuclear Equations

In a nuclear equation, mass numbers and atomic numbers are balanced instead of elements.

The example here to the right depicts a radioactive decay; specifically an alpha decay.

The helium ion is called an alpha particle.

In a nuclear equation, mass numbers and atomic numbers are balanced instead of elements.

The example here to the right depicts a radioactive decay; specifically an alpha decay.

The helium ion is called an alpha particle.

3 Common types of Radioactive Decay

Alpha decay

Beta decay - a ß- particle is a subatomic nuclear particle essentially equivalent to an electron and a ß+ particle is a positively charge electron, called a positron.

Gamma decay - high energy photons are emitted which have virtually no mass nor charge.

Alpha decay

Beta decay - a ß- particle is a subatomic nuclear particle essentially equivalent to an electron and a ß+ particle is a positively charge electron, called a positron.

Gamma decay - high energy photons are emitted which have virtually no mass nor charge.

Nuclear electrons?

Modern theory has shown that a neutron is actually comprised of a proton and an electron.

So, if a nucleus emits an electron, it has really transformed a neutron into a proton.

Also, if a nucleus absorbs an electron, it will convert a proton into a neutron.

Modern theory has shown that a neutron is actually comprised of a proton and an electron.

So, if a nucleus emits an electron, it has really transformed a neutron into a proton.

Also, if a nucleus absorbs an electron, it will convert a proton into a neutron.

Common Particles in Nuclear Reactions

Neutrons (10n)

Protons (11p or 1

1H) Electrons (0

-1e) Alpha Particles (4

2He or 42)

Beta- Particles (0-1e or 0

-1) Gamma (0

0) - Gamma radiation consists of high-energy photons, with a mass far too little for consideration.

Positron (01e) - A positron is a positively

charged electron. It has the mass of an electron but a positive charge.

Neutrons (10n)

Protons (11p or 1

1H) Electrons (0

-1e) Alpha Particles (4

2He or 42)

Beta- Particles (0-1e or 0

-1) Gamma (0

0) - Gamma radiation consists of high-energy photons, with a mass far too little for consideration.

Positron (01e) - A positron is a positively

charged electron. It has the mass of an electron but a positive charge.

Differentiating the Radiations

Alpha emissions are the heaviest and thus have the least penetrating power.

Beta emissions have masses much smaller than protons or neutrons, so they have more penetrating power. In terms of penetrating power, =100*.

Gamma emissions have essentially no mass, so they are the most powerful. In terms of penetrating power. = 100*.

Alpha emissions are the heaviest and thus have the least penetrating power.

Beta emissions have masses much smaller than protons or neutrons, so they have more penetrating power. In terms of penetrating power, =100*.

Gamma emissions have essentially no mass, so they are the most powerful. In terms of penetrating power. = 100*.

Try this

Write a nuclear equation for the process when mercury-201 undergoes electron capture.

Write a nuclear equation for the process when mercury-201 undergoes electron capture.

To answer this question:

First we have to understand what mercury-201 is. Since mercury is always atomic number 80, this isotope is 201

80Hg. Since we are capturing an electron, the electron must be a reactant.

Now we add up mass numbers and atomic numbers. (201 + 0 = 201 and 80 + -1 =79).

Element 79 is gold, so the answer is: 201

80Hg + 0-1e -->201

79Au

First we have to understand what mercury-201 is. Since mercury is always atomic number 80, this isotope is 201

80Hg. Since we are capturing an electron, the electron must be a reactant.

Now we add up mass numbers and atomic numbers. (201 + 0 = 201 and 80 + -1 =79).

Element 79 is gold, so the answer is: 201

80Hg + 0-1e -->201

79Au

Try another

Thorium-231 decays into protactinium-231. What is the balanced equation? What other particle(s) is/are involved in the reaction?

Thorium-231 decays into protactinium-231. What is the balanced equation? What other particle(s) is/are involved in the reaction?

The answers are:

23190Th --> 231

91Pa + 0-1e

The extra particle is an electron, but because it is being emitted, it would be called a Beta emission.

23190Th --> 231

91Pa + 0-1e

The extra particle is an electron, but because it is being emitted, it would be called a Beta emission.

Nuclear Transformations

The first manmade conversion of one nucleus into another was performed by Sir Ernest Rutherford (1919).

Rutherford bombarded a nitrogen-14 atom with alpha particles to produce an oxygen-17 atom plus a proton. 14

7N + 42He --> 17

8O + 11H

The shorthand version of this reaction is 14

7N(,p)178O WHY???

The first manmade conversion of one nucleus into another was performed by Sir Ernest Rutherford (1919).

Rutherford bombarded a nitrogen-14 atom with alpha particles to produce an oxygen-17 atom plus a proton. 14

7N + 42He --> 17

8O + 11H

The shorthand version of this reaction is 14

7N(,p)178O WHY???

Now try this one:

Write the balanced nuclear equation for the process noted by the shorthand:

2713Al(n,)24

11Na

Write the balanced nuclear equation for the process noted by the shorthand:

2713Al(n,)24

11Na

Now try this one:

Write the balanced nuclear equation for the process noted by the shorthand:

2713Al(n,)24

11Na

2713Al + 1

0n --> 2411Na + 4

2He

Write the balanced nuclear equation for the process noted by the shorthand:

2713Al(n,)24

11Na

2713Al + 1

0n --> 2411Na + 4

2He

Nuclear Stability

Why are some nuclei more stable than others?

To be honest, there are several factors, most of which are beyond the scope of this course.

However, there are a few easy to see indications of nuclear stability.

Why are some nuclei more stable than others?

To be honest, there are several factors, most of which are beyond the scope of this course.

However, there are a few easy to see indications of nuclear stability.

Did you ever wonder…?

We know that like charges repel each other, yet a nucleus can have dozens of positively charged protons held together. Why?

Neutrons are a major reason. All nuclei with 2 or more protons have neutrons. The neutrons and the protons meld by a force of nature, different than gravity or electromagnetism, called the strong (nuclear) force.

Because of the way this force binds the protons and neutrons together, the ratio of protons to neutrons is an issue.

We know that like charges repel each other, yet a nucleus can have dozens of positively charged protons held together. Why?

Neutrons are a major reason. All nuclei with 2 or more protons have neutrons. The neutrons and the protons meld by a force of nature, different than gravity or electromagnetism, called the strong (nuclear) force.

Because of the way this force binds the protons and neutrons together, the ratio of protons to neutrons is an issue.

Smaller Atoms vs Bigger Atoms

In smaller atoms, most stable atoms have neutron to proton ratios of about 1.00.

As isotopes increase in atomic number, most stable isotopes have increasingly larger ratios of neutrons to protons.

To our knowledge, any isotope with an atomic number greater than or equal to 84 would be radioactive.

In smaller atoms, most stable atoms have neutron to proton ratios of about 1.00.

As isotopes increase in atomic number, most stable isotopes have increasingly larger ratios of neutrons to protons.

To our knowledge, any isotope with an atomic number greater than or equal to 84 would be radioactive.

Some stability trends

Of the 265 known stable isotopes: 157 of them have even numbers of protons and neutrons.

53 of them have an even number of protons but an odd number of neutrons.

50 of them have an odd number of protons but an even number of neutrons.

Only 5 of them have odd numbers of both protons and neutrons.

Of the 265 known stable isotopes: 157 of them have even numbers of protons and neutrons.

53 of them have an even number of protons but an odd number of neutrons.

50 of them have an odd number of protons but an even number of neutrons.

Only 5 of them have odd numbers of both protons and neutrons.

Magic Numbers

For some reason, nuclei with 2,8,20,28,50 or 82 protons and/or 2,8,20,28,50,82, or 126 neutrons are generally more stable than isotopes without these numbers.

When we think of substances that shield radiation, we tend to think of lead. The most common isotope of lead is 208

82Pb; that means it has 82 protons and 126 neutrons.

For some reason, nuclei with 2,8,20,28,50 or 82 protons and/or 2,8,20,28,50,82, or 126 neutrons are generally more stable than isotopes without these numbers.

When we think of substances that shield radiation, we tend to think of lead. The most common isotope of lead is 208

82Pb; that means it has 82 protons and 126 neutrons.

Decays and Half-lifes

When a radioactive substance decays, the amount of that particular isotope will decrease.

We call the rate of decay the half-life, because it is the time needed for exactly 1/2 of the isotope to decay.

When a radioactive substance decays, the amount of that particular isotope will decrease.

We call the rate of decay the half-life, because it is the time needed for exactly 1/2 of the isotope to decay.

More on Half-life

If we examine the graph to the right, we see that we started with 50 g of the isotope. Each subsequent point represents half of the previous mass (50 to 25 to 12.5 to 6.25 to 3.125 to 1.5625 to .78125).

Each point is approximately 24 days apart; The half-life for this substance is 24 days.

If we examine the graph to the right, we see that we started with 50 g of the isotope. Each subsequent point represents half of the previous mass (50 to 25 to 12.5 to 6.25 to 3.125 to 1.5625 to .78125).

Each point is approximately 24 days apart; The half-life for this substance is 24 days.

Calculations with half-life

Although it is possible to determine the amount remaining of a radioisotope using natural logs { ln(Nt/N0)=-kt } we do not need to do this.

We only work with whole number increments of the half-life.

Although it is possible to determine the amount remaining of a radioisotope using natural logs { ln(Nt/N0)=-kt } we do not need to do this.

We only work with whole number increments of the half-life.

For example

The half-life of an isotope is 8 days. If we start with 100 grams of the isotope, how much is present in 32 days? 32 days/(8 days/half-life) = 4 half-lives. Each half-life divides the previous mass in half.

100g/2 = 50g/2 = 25g/2 = 12.5g/2 = 6.25g There would be 6.25 g of that isotope left.

The half-life of an isotope is 8 days. If we start with 100 grams of the isotope, how much is present in 32 days? 32 days/(8 days/half-life) = 4 half-lives. Each half-life divides the previous mass in half.

100g/2 = 50g/2 = 25g/2 = 12.5g/2 = 6.25g There would be 6.25 g of that isotope left.

You try one

The half-life of Bismuth-211 is 185 years. How much time would it take for a 360 g sample to decay to 11.25 g?

The half-life of Bismuth-211 is 185 years. How much time would it take for a 360 g sample to decay to 11.25 g?

You try one

The half-life of Bismuth-211 is 185 years. How much time would it take for a 360 g sample to decay to 11.25 g?

360g/2=180g/2=90g/2=45g/2=22.5g/2=11.25g.

That is 5 half-lives. 5 half-lives*185 years/half-life = 925 years.

The half-life of Bismuth-211 is 185 years. How much time would it take for a 360 g sample to decay to 11.25 g?

360g/2=180g/2=90g/2=45g/2=22.5g/2=11.25g.

That is 5 half-lives. 5 half-lives*185 years/half-life = 925 years.

Fusion vs. Fission

Fission and Fusion are two types of highly exothermic nuclear reactions, different than the decays covered earlier.

Fusion means to bring two smaller nuclei together to make a larger nucleus. 13

6C + 136C --> 25

11Na + 11p

Fission means to break a larger nucleus into 2 or more smaller nuclei. 235

92U + 10n --> 137

52Te + 9740Zr + 2 1

0n

Fission and Fusion are two types of highly exothermic nuclear reactions, different than the decays covered earlier.

Fusion means to bring two smaller nuclei together to make a larger nucleus. 13

6C + 136C --> 25

11Na + 11p

Fission means to break a larger nucleus into 2 or more smaller nuclei. 235

92U + 10n --> 137

52Te + 9740Zr + 2 1

0n

How much Energy are we talking about?

In fusion and fission, a tiny, almost meaningless mass of each affected nucleus is converted to energy.

Einstein theorized that the amount of energy was dependent on the mass lost and the square of the speed of light. E = mc2.

In fusion and fission, a tiny, almost meaningless mass of each affected nucleus is converted to energy.

Einstein theorized that the amount of energy was dependent on the mass lost and the square of the speed of light. E = mc2.

So, how much energy is that?

Well if each uranium atom in a given fission process loses the mass of one electron (9.11x10-31 kg):

E = mc2

E = (9.11x10-31 kg) *(3.0x108 m/s)2

E = 8.2x10-14 J

Well if each uranium atom in a given fission process loses the mass of one electron (9.11x10-31 kg):

E = mc2

E = (9.11x10-31 kg) *(3.0x108 m/s)2

E = 8.2x10-14 J

But that seems like a small number!

8.2x10-14 J is a small amount, but that was for just one atom or uranium. If we had 1 mg of uranium (about the mass of a cystal of salt), that would contain roughly 2.5 x 1018 atoms of uranium.

8.2x10-14 J/atom * 2.5 x 1018 atoms =2.1x105 J

That is almost enough energy to handle the electrical needs of this school for a day - from a tiny starting mass.

8.2x10-14 J is a small amount, but that was for just one atom or uranium. If we had 1 mg of uranium (about the mass of a cystal of salt), that would contain roughly 2.5 x 1018 atoms of uranium.

8.2x10-14 J/atom * 2.5 x 1018 atoms =2.1x105 J

That is almost enough energy to handle the electrical needs of this school for a day - from a tiny starting mass.

Think about this

If we could convert the .25 kg mass of a banana peel (using our Mr. Fusion power supply) into pure energy,

E = mc2

E = (.25 kg)*(3.0x108 m/s)2

E = 2.25x1016 J That’s enough energy to run New York City for a year (with enough energy left over to go Back to the Future)!

If we could convert the .25 kg mass of a banana peel (using our Mr. Fusion power supply) into pure energy,

E = mc2

E = (.25 kg)*(3.0x108 m/s)2

E = 2.25x1016 J That’s enough energy to run New York City for a year (with enough energy left over to go Back to the Future)!