When ²10}₈2}Pb decays to ²06}₈0}Pb, it emits

Correct Answer is : an alpha particle

two alpha and two beta particles

one alpha and one beta particle

Correct Answer is : an alpha particle

Test Index

JAMB Physics 2014 Paper

Question : 39

Total: 41

When

^{210}_{82}Pb

^{206}_{80}Pb

, it emits

Solution:

Concept:

Alpha decay reduces the mass number by 4 and the atomic number by 2.

Explanation:

Initial nucleus:

^{210}_{82}\mathrm{Pb}

; final nucleus:

^{206}_{80}\mathrm{Pb}

Mass number change:

210 - 206 = 4

Atomic number change:

82 - 80 = 2

An alpha particle,

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

, has mass number 4 and atomic number 2, so only one alpha particle is emitted:

^{210}_{82}\mathrm{Pb} \rightarrow ^{206}_{80}\mathrm{Pb} + ^{4}_{2}\mathrm{He}

Answer:

an alpha particle.

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