Calculate the mass number and atomic number of the final daughter nuclide in the given rad… — JAMB UTME

Calculate the mass number and atomic number of the final daughter nuclide in the given radioactive decay series.

Answer:When a nucleus of Thorium-232 (

^{232}_{90}\text{Th}

) undergoes a decay series emitting

6

alpha particles (

\alpha

) and

4

beta particles (

\beta^-

), the resulting stable daughter nuclide has a mass number of 【208】 and an atomic number of 【82】.

The final stable daughter nuclide has a mass number of 208 and an atomic number of 82.

Emission of 6 alpha particles reduces the mass number by

6 \times 4 = 24

(from 232 to 208) and the atomic number by

6 \times 2 = 12

. The subsequent emission of 4 beta particles increases the atomic number by

4 \times 1 = 4

, resulting in a final atomic number of

90 - 12 + 4 = 82

Calculate the total change in mass number (

A

) caused by alpha and beta particle emissions.

Each alpha particle (

^{4}_{2}\text{He}

) reduces

A

by 4, while beta particles (

^{0}_{-1}\text{e}

) do not change

A

. Total decrease in

A = 6 \times 4 = 24

Alpha particles carry 4 atomic mass units, whereas beta particles have virtually zero mass number.

Determine the final mass number.

Final mass number =

232 - 24 = 208

Subtracting the lost nucleon mass from the initial mass number gives the mass number of the resulting nucleus.

Calculate the total change in atomic number (

Z

) caused by alpha and beta particle emissions.

Each alpha particle reduces

Z

by 2, while each beta particle increases

Z

by 1. Total change in

Z = (6 \times -2) + (4 \times +1) = -12 + 4 = -8

Alpha decay removes 2 protons, whereas beta minus decay converts a neutron into a proton, increasing nuclear charge by 1.

Determine the final atomic number.

Final atomic number =

90 - 8 = 82

Applying the net charge change to the initial atomic number of Thorium (90) yields 82, which corresponds to Lead (Pb).

Conservation of mass number and atomic number in decay series

Question