The atomic mass number of the element thorium | vedclass

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(B) Let the number of alpha particles emitted be $n_{\alpha}$ and the number of beta particles emitted be $n_{\beta}$.The change in mass number is due

Vedclass · Accepted Answer

$\alpha = 6, \beta = 4$

Vedclass · Answer

(B) Let the number of alpha particles emitted be $n_{\alpha}$ and the number of beta particles emitted be $n_{\beta}$.The change in mass number is due to alpha particles only,as each alpha particle reduces the mass number by $4$.$n_{\alpha} = \frac{A - A'}{4} = \frac{232 - 208}{4} = \frac{24}{4} = 6$.The change in atomic number is given by $Z' = Z - 2n_{\alpha} + n_{\beta}$.Substituting the values: $82 = 90 - 2(6) + n_{\beta}$.$82 = 90 - 12 + n_{\beta}$.$82 = 78 + n_{\beta}$.$n_{\beta} = 82 - 78 = 4$.Therefore,$6$ alpha particles and $4$ beta particles are emitted.

The atomic mass number of the element thorium is $232$ and its atomic number is $90$. The end product of this radioactive element is an isotope of lead (atomic mass $208$ and atomic number $82$). The number of alpha and beta particles emitted is:

Similar Questions

The atomic weight of thorium is $A = 232$ and the atomic number is $Z = 90$. After disintegration,the final element obtained is $Pb$,which has $A = 208$ and $Z = 82$. In this process,the number of emitted $\alpha$ and $\beta$ particles are:

An isotope of the original nucleus can be formed in a radioactive decay,with the emission of which of the following particles?

Alpha particles are

$\beta$-decay means emission of electron from

The nucleus $_{90}Th^{228}$ transforms into $_{83}Bi^{212}$. How many $\alpha$-particles and $\beta$-particles are emitted during this process?

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