Nuclear Equations (WJEC GCSE Physics)

Revision Note

Author

Katie M

Expertise

Physics

Balancing Nuclear Equations

Radioactive decay events can be shown using a decay equation
A decay equation is similar to a chemical reaction equation, which means:
- The particles present before the decay are shown before the arrow
- The particles produced in the decay are shown after the arrow
- The sum of the mass and atomic numbers before the reaction must be the same as the sum of the mass and atomic numbers after the reaction

The following decay equation shows polonium-212 undergoing alpha decay

${}_{84}^{212}{Po} \to {}_{82}^{208}{Pb} + {}_{2}^{4}α$

The polonium nucleus emits an alpha particle (i.e. a helium nucleus)
This causes both its proton and nucleon numbers to decrease, so it changes into a new element, lead-208

Alpha Decay Equation

When the alpha particle is emitted from the unstable nucleus, the nucleon number and proton number of the nucleus changes
- The mass number decreases by 4
- The atomic number decreases by 2

Representing Alpha Decay

When a nucleus decays by emitting an alpha particle, both the proton (atomic) and nucleon (mass) numbers decrease

Beta Decay Equation

During beta decay, an electron is emitted
Meanwhile, the proton number increases by one, and the nucleon number stays the same

Representing Beta Decay

When a nucleus decays by emitting a beta particle, the proton (atomic) number increases while the nucleon (mass) number stays the same

Gamma Decay

During gamma decay, a gamma ray is emitted from an unstable nucleus
The emitted gamma ray has a lot of energy, but no mass or charge
The process that makes the nucleus less energetic but does not change its structure

Representing Gamma Decay

When a nucleus decays by emitting a gamma ray, both the proton (atomic) and nucleon (mass) numbers stay the same

Worked example

Polonium-210 undergoes alpha decay. It forms lead, which has the element symbol Pb.

${}_{84}^{210}{Po} \to {}_{y}^{x}{Pb} + α$

What are the nucleon and proton numbers of the lead isotope?

Worked Example Alpha Decay, downloadable IGCSE & GCSE Physics revision notes

ANSWER: A

Step 1: Recall the nucleon and proton numbers of alpha

The emitted alpha particle is made of two protons and two neutrons
The proton number is 2
The nucleon number is 4

${}_{84}^{210}{Po} \to {}_{y}^{x}{Pb} + {}_{2}^{4}α$

Step 2: Calculate the proton number of lead

The proton numbers on both sides must add up to the same number
So, the new proton number will decrease by 2

y = 84 – 2 = 82

The lead nucleus has a proton number of 82

Step 3: Calculate the nucleon number of the lead isotope

The nucleon numbers on both sides must add up to the same number
So, the new nucleon number will decrease by 4

x = 210 – 4 = 206

The lead isotope has a mass number of 206

Worked example

Sodium-24 undergoes beta decay. It forms magnesium, which has the element symbol Mg.

${}_{11}^{24}{Na} \to {}_{y}^{x}{Mg} + β$

What are the nucleon and proton numbers of the magnesium isotope?

Worked example beta decay, downloadable IGCSE & GCSE Physics revision notes

ANSWER: D

Step 1: Recall the changes that happen during beta decay

During beta decay, the nucleon number remains the same and the proton number increases by 1
The proton number can be written −1
The nucleon number can be written 0

${}_{11}^{24}{Na} \to {}_{y}^{x}{Mg} + {}_{- 1}^{0}β$

Step 2: Calculate the proton number of magnesium

The proton numbers on both sides must add up to the same number
So, the new proton number will increase by 1

y = 11 + 1 = 12

The magnesium nucleus has a proton number of 12

Step 3: Calculate the nucleon number of the magnesium isotope

The nucleon numbers on both sides must add up to the same number
So, the new nucleon number will stay the same

x = 24 + 0 = 24

The magnesium isotope has a nucleon number of 24

Exam Tip

You are not expected to know the names of the elements produced during radioactive decay, but you do need to be able to calculate the nucleon and proton numbers by making sure they are balanced on either side of the reaction.

Writing Nuclear Equations

Higher

Given some information about different nuclei, you could be asked to write full, balanced nuclear equations involving alpha, beta and gamma decay
Remember the key rules for before and after the reaction:
- The total nucleon number must be the same
- The total proton number must be the same

The worked examples below illustrate the kinds of questions that could be asked

Worked example

A nucleus of iodine-131 $({}_{53}^{131}I)$ decays into xenon $(Xe)$ by emitting beta and gamma radiation.

Write a balanced equation for this decay.

Answer:

Step 1: Recall the symbols for beta and gamma particles

A beta particle is an electron, which changes proton number by 1 while nucleon number stays the same
So, the symbol for beta is: ${}_{- 1}^{0}β$ or ${}_{- 1}^{0}e$

Gamma radiation is an electromagnetic wave which does not cause any changes to proton or nucleon number
The symbol for gamma is: $γ$

Step 2: Use information from the question to write a word equation

Underline or highlight the keywords in the question

"A nucleus of iodine-131 decays into xenon by emitting beta and gamma radiation"

So, as a word equation, this can be written

iodine-131 → xenon + beta + gamma

Step 3: Write the balanced nuclear equation using the correct symbols

In ${}_{Z}^{A}X$ notation, iodine-131 is ${}_{53}^{131}I$ , and we can write xenon as ${}_{Z}^{A}{Xe}$ for now
The nuclear equation can be written

${}_{53}^{131}I \to {}_{Z}^{A}{Xe} + {}_{- 1}^{0}β + γ$

The total proton numbers and total nucleon numbers must be equal on both sides
- Total nucleon number: 131 = A + 0 + 0
- Nucleon number of xenon: A = 131
- Total proton number: 53 = Z − 1 + 0
- Proton number of xenon: Z = 53 + 1 = 54

The balanced equation is:

${}_{53}^{131}I \to {}_{54}^{131}{Xe} + {}_{- 1}^{0}β + γ$

Worked example

A nucleus decays by beta particle emission to form polonium-210. Polonium-210 then undergoes alpha decay to form a stable element.

The table below shows some of the elements near polonium in the periodic table.

Element	Lead	Bismuth	Polonium	Astatine	Radon
Symbol	Pb	Bi	Po	At	Rn
Proton number	82	83	84	85	86

Write two balanced nuclear equations to represent the decays.

$. . . . . . . . \to . . . . . . . . + . . . . . . . .$

Answer:

Step 1: Recall the symbols for alpha and beta particles

An alpha particle is a helium nucleus, which contains 2 protons and 2 neutrons
So, the symbol for alpha is: ${}_{2}^{4}α$ or ${}_{2}^{4}{He}$

A beta particle is an electron, which changes proton number by 1 while nucleon number stays the same
So, the symbol for beta is: ${}_{- 1}^{0}β$ or ${}_{- 1}^{0}e$

Step 2: Use information from the question to write the first nuclear equation

Underline or highlight the keywords in the question

"A nucleus decays by beta particle emission to form polonium-210"

We need to work out what the nucleus is, but for now, we can call it nucleus X
So, as a word equation, this can be written

nucleus X → polonium-210 + beta

Polonium-210 has a nucleon number of 210, and from the table, it has a proton number of 84
In ${}_{Z}^{A}X$ notation, polonium-210 is ${}_{84}^{210}{Po}$
The first nuclear equation is:

${}_{b}^{a}X \to {}_{84}^{210}{Po} + {}_{- 1}^{0}β$

Step 3: Balance the first nuclear equation and determine the missing nucleus:

The total proton numbers and total nucleon numbers must be equal on both sides
- Total proton number: b = 84 − 1 = 83
- Total nucleon number: a = 210 + 0 = 210

Therefore, from the table, nucleus X is bismuth-210, which is ${}_{83}^{210}{Bi}$
So, the first balanced nuclear equation is

${}_{83}^{210}{Bi} \to {}_{84}^{210}{Po} + {}_{- 1}^{0}β$

Step 4: Use information from the question to write the second nuclear equation

Underline or highlight the keywords in the question

"Polonium-210 then undergoes alpha decay to form a stable element"

We need to work out what the stable element is, but for now, we can call it nucleus Y
So, as a word equation, this can be written

polonium-210 → nucleus Y + alpha

The second nuclear equation is:

${}_{84}^{210}{Po} \to {}_{d}^{c}Y + {}_{2}^{4}α$

Step 5: Balance the second nuclear equation and determine the missing nucleus:

The total proton numbers and total nucleon numbers must be equal on both sides
- Total proton number: 84 = d + 2
- So, proton number of Y: d = 84 − 2 = 82
- Total nucleon number: 210 = c + 4
- So, nucleon number of Y: c = 210 − 4 = 206

Therefore, from the table, nucleus Y is lead-206, which is ${}_{82}^{206}{Pb}$
So, the second balanced nuclear equation is

${}_{84}^{210}{Po} \to {}_{82}^{206}{Pb} + {}_{2}^{4}α$

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Nuclear Equations (WJEC GCSE Physics)

Revision Note

Alpha Decay Equation

Representing Alpha Decay

Beta Decay Equation

Representing Beta Decay

Gamma Decay

Representing Gamma Decay

Higher

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Author: Katie M