Question

After the emission of a $\beta$-particle followed by $\alpha$-particle from ${}_{83}^{214}Bi$, the number of neutrons in the atom is:

• A. 130
• B. 129
• C. 128
• D. 127

Answer 1

The correct option is C 128

When a beta particle is lost, the atomic number increases by 1 and the mass number remains unchanged. When an alpha particle is lost, the atomic number decreases by 2 and the mass number decreases by 4.

The decay equation is given below.

${}_{83}^{214}Bi{\stackrel{\beta emission}{\to }}_{84}^{214}Po{\stackrel{\alpha emission}{\to }}_{82}^{210}Pb$

The number of neutrons in ${}_{82}^{210}Pb$ is $210-82=128$.