A ²³²₉₀Th nucleus ultimately changes into ²⁰⁸₈₂Pb. During the complete transformation only α-decay and β⁻-decay occur. Let n(α) be the number of α-particles emitted and n(β) the number of β⁻-particles emitted. If N = n(α)+n(β), find N.
Physics · Nuclear physics · radioactivity and law of decay
JEE Mains NEET JEE Advanced

Question

A ²³²₉₀Th nucleus ultimately changes into ²⁰⁸₈₂Pb. During the complete transformation only α-decay and β⁻-decay occur. Let n(α) be the number of α-particles emitted and n(β) the number of β⁻-particles emitted. If N = n(α)+n(β), find N.

Physics · Nuclear physics · radioactivity and law of decay

G1 Physics · Accepted Answer

First use the mass-number change. Each α-particle reduces the mass number by 4, whereas β⁻-decay does not change the mass number. 232−4n(α) = 208 4n(α) = 24 n(α) = 6. Now track atomic number. After six α-decays: Z = 90−2×6 = 78. The final atomic number is 82. Each β⁻-decay increases Z by 1, so 78+n(β) = 82. Therefore, n(β) = 4. Hence, N = 6+4 = 10.