Question

${}_5{B}^{10}$ + ${}_{2}H{e}^4$ $\to$ X + ${}_0{n}^1$

'X' in the above reaction is

• A. ₇N¹²
• B. ₇N¹³
• C. ₆B¹³
• D. ₆B¹⁴

Answer 1

The correct option is B

₇N¹³

${}_5{B}^{10}$ + ${}_{2}H{e}^4$ $\to$ X + ${}_0{n}^1$

^{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,

5+2=y+0

=> y= 7

Hence X is nitrogen

Similarly, the mass number will also be conserved:

Therefore,

10+4=z+1

=>z=13

So X is ${}_7{N}^{13}$