Physics - Modern Physics Question with Solution | TestHub

For a nuclear reaction to be possible, both the mass number ($A$) and the atomic number ($Z$) must be conserved.

Checking the given options:

For ${}_5{\text{B}}^{10}+{}_2{\text{He}}^4\to {}_7{\text{N}}^{13}+{}_1{\text{H}}^1$:
LHS: $Z=5+2=7$
RHS: $Z=7+1=8$
Since $7\ne 8$, the atomic number is not conserved.

For ${}_{11}{\text{Na}}^{23}+{}_1{\text{H}}^1\to {}_{10}{\text{N}}^{20}+{}_2{\text{He}}^4$:
LHS: $Z=11+1=12$, $A=23+1=24$
RHS: $Z=10+2=12$, $A=20+4=24$
Although $Z$ and $A$ are mathematically balanced, the chemical symbol for $Z=10$ is Neon ($\text{Ne}$), not Nitrogen ($\text{N}$).

For ${}_{93}{\text{Np}}^{239}\to {}_{94}{\text{Pu}}^{239}+{}_{-1}{\text{e}}^0$:
LHS: $Z=93$, $A=239$
RHS: $Z=94-1=93$, $A=239+0=239$
Both the atomic number and mass number are conserved. This is a valid ${\beta }^{-}$ decay reaction.

For ${}_7{\text{N}}^{11}+{}_1{\text{H}}^1\to {}_6{\text{C}}^{12}+{}_{-1}{\text{e}}^0$:
LHS: $Z=7+1=8$
RHS: $Z=6-1=5$
Since $8\ne 5$, the atomic number is not conserved.

Answer: ${}_{93}{\text{Np}}^{239}\to {}_{94}{\text{Pu}}^{239}+{}_{-1}{\text{e}}^0$