A radioactive parent nucleus of Thorium, ^{232}_{90}\text{Th}, undergoes a natural radioac… — JAMB UTME

A radioactive parent nucleus of Thorium, $^{232}_{90}\text{Th}$ , undergoes a natural radioactive decay series by emitting a total of $6$ $\alpha$ -particles and $4$ $\beta^-$ -particles to form a stable daughter isotope of Lead ( $\text{Pb}$ ). Calculate the number of neutrons present in the nucleus of the resulting daughter isotope.

Answer: 126

Answer

The resulting daughter nucleus contains 126 neutrons.

Emitting

6

\alpha

-particles reduces the mass number by

6 \times 4 = 24

units and the atomic number by

6 \times 2 = 12

units. Emitting

4

\beta^-

-particles leaves the mass number unchanged while increasing the atomic number by

4 \times 1 = 4

units. Consequently, the daughter nucleus has a mass number

A = 232 - 24 = 208

and an atomic number

Z = 90 - 12 + 4 = 82

. The number of neutrons is

N = A - Z = 208 - 82 = 126

Step-by-Step Solution

Calculate the mass number (

A

) of the daughter nucleus after all emissions

A = 232 - (6 \times 4) = 208

An alpha particle carries away 4 mass units (

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

), while a beta-minus particle carries 0 mass units (

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

Calculate the atomic number (

Z

) of the daughter nucleus after all emissions

Z = 90 - (6 \times 2) + (4 \times 1) = 82

Each alpha decay reduces nuclear charge by 2, and each beta-minus decay increases nuclear charge by 1 due to neutron-to-proton conversion.

Compute the number of neutrons (

N

)

N = A - Z = 208 - 82 = 126

The number of neutrons in any nuclide is given by subtracting the atomic number (protons) from the mass number (nucleons).

Key Concept

Mass and Atomic Number Conservation in Radioactive Decay Chains

Estimated Time:2m 0s

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