Question

On the bombardment of ${}_7{N}^{14}$ with alpha particles, the nucleus of the product after the release of a proton will be

• A. ₈O¹⁷
• B. ₈O¹⁸
• C. ₉F¹⁷
• D. ₉F¹⁸

Answer 1

The correct option is B

₈O¹⁸

${}_7{N}^{14}$ + ${}_{2}H{e}^4$ $\to$ X + ${}_1{p}^0$

^{In the above reaction let y be the number of protons of X and z be the mass number of X.}

^{Therefore X can be represented as: ${}_y{X}^z$}

Since total number of protons before and after the reaction will remain same:

Therefore,

7+2=y+1

=> y= 8

Hence X is oxygen

Similarly, the mass number will also be conserved:

Therefore,

14+4=z+0

=>z=18

So X is ${}_8{O}^{18}$