Question

An atomic nucleus ${\rm A}$ is composed of $84$ protons and $128$ neutrons, 

(i) The nucleus ${\rm A}$ emits an alpha particle and is transformed into nucleus ${\rm B}$. What is the composition of the nucleus ${\rm B}$? 

(ii) The nucleus ${\rm B}$ emits a beta particle and is transformed into a nucleus $C$. What is the composition of the nucleus $C$? 

(iii) Does the composition of nucleus $C$ change if it emits gamma radiation?

Answer 1

Hint: When an atom is subjected to the alpha decay, the nucleus tends to the reduction of mass number by $4$ and reduction of atomic number by $2$. Then, that atom is subjected to beta decay, the atomic number gains by $1$, and the mass number remains the same. When an atom is subjected to gamma decay, there will not be any change in the atomic and mass number. In general, gamma decay normally occurs during both alpha decay as well as in beta decay.

During alpha emission, the alpha particles with the atomic number $2$ and the mass number $4$ will emit.

During beta emission, the beta particles with a $ - 1$ atomic number and zero mass number will emit.

During gamma emission, there will not be any change in the atomic and mass number.

Complete step by step solution:

(i) Given, the nucleus of an atom contains $84$ protons and $128$neutrons. Hence, the atomic number of the atomic nucleus is the number of protons in the nucleus and the mass number is the sum of the protons and neutrons present in it.

Hence,

An atomic number of atomic nuclei ${\rm A}$ is $84$.

Mass number of the atomic nucleus ${\rm A}$ is $84 + 128 = 212$.

Thus, the atomic nucleus ${\rm A}$ is denoted by ${}_{84}^{212}{\rm A}$.

The atomic nucleus $A$ subjected to alpha emission,

${}_{84}^{212}{\rm A}\xrightarrow{{ - \alpha }}{}_2^4\alpha  + {}_{82}^{208}{\rm B}$

Hence, the composition of the atomic nucleus ${\rm B}$ is ${}_{82}^{208}{\rm B}$.

(ii) The atomic nucleus ${\rm B}$ is subjected to the beta emission, thus the atomic number of the nucleus will get increase by one,

${}_{82}^{208}{\rm B}\xrightarrow{{ - \beta }}{}_{ - 1}^0\beta  + {}_{83}^{208}C$

Hence, the composition of the atomic nucleus $C$ is ${}_{83}^{208}C$.

(iii) The atomic nucleus $C$ is subjected to the gamma emission, thus there will not be any change in the atomic and mass number. The gamma emission is generally occurring during the alpha and beta emission.

$\therefore$ The atomic nucleus ${\rm A}$ is denoted by ${}_{84}^{212}{\rm A}$.

The composition of the atomic nucleus ${\rm B}$ is ${}_{82}^{208}{\rm B}$.

The composition of the atomic nucleus $C$ is ${}_{83}^{208}C$.

Note:
During the alpha emission of the atomic nucleus the alpha particle generally ${}_2^4He$ will emit from the atomic nucleus. And during the beta emission of the atomic nucleus, the beta particle generally called negatron ${}_{ - 1}^0\beta $ will emit from the atom. And the gamma emission is a common thing that occurs during the alpha and beta emission, it won’t affect the atomic and mass number.