Nuclear Equations (WJEC GCSE Physics): Revision Note

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

${}_{84}{}^{210}\mathrm{Po}\to {}_y{}^x\mathrm{Pb}+\mathrm{\alpha }$

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

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

${}_{84}{}^{210}\mathrm{Po}\to {}_y{}^x\mathrm{Pb}+{}_2{}^4\mathrm{\alpha }$

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

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

${}_{11}{}^{24}\mathrm{Na}\to {}_y{}^x\mathrm{Mg}+\mathrm{\beta }$

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

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

${}_{11}{}^{24}\mathrm{Na}\to {}_y{}^x\mathrm{Mg}+{}_{-1}{}^0\mathrm{\beta }$

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

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.

A nucleus of iodine-131 $\left({}_{53}{}^{131}\mathrm{I}\right)$ decays into xenon $\left(\mathrm{Xe}\right)$ 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: ${}_{-1}{}^0\mathrm{\beta }$  or  ${}_{-1}{}^0\mathrm{e}$

• Gamma radiation is an electromagnetic wave which does not cause any changes to proton or nucleon number

• The symbol for gamma is: $\mathrm{\gamma }$ 

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 ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$ notation, iodine-131 is ${}_{53}{}^{131}\mathrm{I}$, and we can write xenon as ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{Xe}$ for now

• The nuclear equation can be written

${}_{53}{}^{131}\mathrm{I}\to {}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{Xe}+{}_{-1}{}^0\mathrm{\beta }+\mathrm{\gamma }$

• 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:

${}_{53}{}^{131}\mathrm{I}\to {}_{54}{}^{131}\mathrm{Xe}+{}_{-1}{}^0\mathrm{\beta }+\mathrm{\gamma }$

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.

Write two balanced nuclear equations to represent the decays.

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

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

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:  ${}_2{}^4\mathrm{\alpha }$  or  ${}_2{}^4\mathrm{He}$

• A beta particle is an electron, which changes proton number by 1 while nucleon number stays the same

• So, the symbol for beta is:  ${}_{-1}{}^0\mathrm{\beta }$  or  ${}_{-1}{}^0\mathrm{e}$

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 ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$ notation, polonium-210 is ${}_{84}{}^{210}\mathrm{Po}$

• The first nuclear equation is:

${}_b{}^a\mathrm{X}\to {}_{84}{}^{210}\mathrm{Po}+\hspace{0.17em}{}_{-1}{}^0\mathrm{\beta }$

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 ${}_{83}{}^{210}\mathrm{Bi}$

• So, the first balanced nuclear equation is

${}_{83}{}^{210}\mathrm{Bi}\to {}_{84}{}^{210}\mathrm{Po}+\hspace{0.17em}{}_{-1}{}^0\mathrm{\beta }$

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:

${}_{84}{}^{210}\mathrm{Po}\to {}_d{}^c\mathrm{Y}+\hspace{0.17em}{}_2{}^4\mathrm{\alpha }$

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 ${}_{82}{}^{206}\mathrm{Pb}$

• So, the second balanced nuclear equation is

${}_{84}{}^{210}\mathrm{Po}\to {}_{82}{}^{206}\mathrm{Pb}+\hspace{0.17em}{}_2{}^4\mathrm{\alpha }$