Question

The value of A in the following reaction is
${}_4{\text{Be}}^9$ + ${}_2{\text{He}}^4$ = ${}_6{\text{C}}^A$ + ${}_0{\text{n}}^1$

• A. $14$
• B. $10$
• C. $12$
• D. $16$

Answer 1

The correct option is C $12$

The reaction producing neutrons by bombarding beryllium with alpha particles is given as
${}_4^9\text{Be}{+}_2^4\text{He}{\to }_6^A\text{C}{+}_0^1\text{n}$
Since, a neutron has atomic number of zero (it has no positive charge) but has a mass number of one (the same mass as the proton). Hence, the total mass (that is, the total number of nuclear particles or nucleons = 13) entering into the reaction must be the same as nucleons leaving the reaction.
Therefore $A=12$.
${}_4^9\text{Be}{+}_2^4\text{He}{\to }_6^{12}\text{C}{+}_0^1\text{n}$
Thus atomic mass is conserved in the reaction.