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Atomic mass number of an element is 232 ...

Atomic mass number of an element is `232` and its atomic number is `90`. The end product of this radiaoctive element is an isotope of lead (atomic mass `208` and atomic number `82`.) The number of `alpha`-and `beta` -particles emitted are.

A

`6,3`

B

`6,4`

C

`5,5`

D

`4,6`

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The correct Answer is:
To solve the problem, we need to determine the number of alpha (α) and beta (β) particles emitted during the radioactive decay of an element with an atomic mass number of 232 and an atomic number of 90, resulting in a lead isotope with an atomic mass of 208 and an atomic number of 82. ### Step-by-Step Solution: 1. **Identify the Initial and Final Values:** - Initial atomic mass number (A_initial) = 232 - Initial atomic number (Z_initial) = 90 - Final atomic mass number (A_final) = 208 - Final atomic number (Z_final) = 82 2. **Calculate the Change in Atomic Mass Number (ΔA):** \[ ΔA = A_{initial} - A_{final} = 232 - 208 = 24 \] 3. **Calculate the Change in Atomic Number (ΔZ):** \[ ΔZ = Z_{initial} - Z_{final} = 90 - 82 = 8 \] 4. **Understand the Effects of Alpha and Beta Emissions:** - Each alpha particle (α) emitted decreases the atomic mass number (A) by 4 and the atomic number (Z) by 2. - Each beta particle (β) emitted does not change the atomic mass number (A) but decreases the atomic number (Z) by 1. 5. **Set Up the Equations:** Let \( x \) be the number of alpha particles emitted and \( y \) be the number of beta particles emitted. - For the change in atomic mass number: \[ 4x + 0y = 24 \quad \text{(1)} \] - For the change in atomic number: \[ -2x - y = -8 \quad \text{(2)} \] 6. **Solve Equation (1):** From equation (1): \[ 4x = 24 \implies x = \frac{24}{4} = 6 \] 7. **Substitute \( x \) into Equation (2):** Substitute \( x = 6 \) into equation (2): \[ -2(6) - y = -8 \] \[ -12 - y = -8 \implies -y = -8 + 12 \implies -y = 4 \implies y = 4 \] 8. **Final Result:** The number of alpha particles emitted \( x = 6 \) and the number of beta particles emitted \( y = 4 \). ### Conclusion: The number of alpha particles emitted is 6, and the number of beta particles emitted is 4.

To solve the problem, we need to determine the number of alpha (α) and beta (β) particles emitted during the radioactive decay of an element with an atomic mass number of 232 and an atomic number of 90, resulting in a lead isotope with an atomic mass of 208 and an atomic number of 82. ### Step-by-Step Solution: 1. **Identify the Initial and Final Values:** - Initial atomic mass number (A_initial) = 232 - Initial atomic number (Z_initial) = 90 - Final atomic mass number (A_final) = 208 ...
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