In a radioactive decay,an element ^A_Z X emit | vedclass

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(A) When a nucleus undergoes radioactive decay,the emission of an $\alpha$-particle $(^4_2 He)$ decreases the atomic number by $2$ and the mass number

Vedclass · Accepted Answer

$Z-5, A-16$

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(A) When a nucleus undergoes radioactive decay,the emission of an $\alpha$-particle $(^4_2 He)$ decreases the atomic number by $2$ and the mass number by $4$.The emission of a $\beta$-particle $(^0_{-1} e)$ increases the atomic number by $1$ and leaves the mass number unchanged.Gamma $(\gamma)$ emission does not change the atomic number or the mass number.Given: $n_{\alpha} = 4$,$n_{\beta} = 3$,$n_{\gamma} = 8$.Final atomic number $Z' = Z - 2(n_{\alpha}) + 1(n_{\beta}) = Z - 2(4) + 3 = Z - 8 + 3 = Z - 5$.Final mass number $A' = A - 4(n_{\alpha}) + 0(n_{\beta}) = A - 4(4) = A - 16$.Therefore,the resulting nucleus has atomic number $Z-5$ and mass number $A-16$.

In a radioactive decay,an element $^A_Z X$ emits four $\alpha$-particles,three $\beta$-particles and eight $\gamma$-photons. The atomic number and mass number of the resulting final nucleus are

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