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Q.
In the nuclear fission reaction, $ _{2}^{4}He + \,_{7}^{14}N \to \,_{p}^{q}x + \,_{1}H^{1} $ the nucleus $ _{p}^{q}X $ is :
J & K CETJ & K CET 2001
Solution:
In order that reaction holds true mass number and atomic number remain conserved.
Given $ _{2}^{4}He + \,_{7}^{14}N \to \,_{p}^{q}x + \,_{1}H^{1} $
Equating mass number $ 14+4=q+1 $
$ \Rightarrow $ $ q=17 $
Equating atomic number $ 7+2=p+1 $
$ \Rightarrow $ $ p=8 $
Hence, unknown element is an isotope of oxygen of mass $17$.