A) nitrogen of mass 16
B) nitrogen of mass 17
C) oxygen of mass 16
D) oxygen of mass 17
Correct Answer: D
Solution :
In order that reaction holds true mass number and atomic number remain conserved. Given\[_{2}^{4}He+_{7}^{14}N\xrightarrow{{}}_{p}^{q}X{{+}_{1}}{{H}^{1}}\] 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.
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