Q22 / 155 · WBJEE 2021 · Chemistry

For the reaction \(14 N+\alpha \longrightarrow{}^{17} O+p\), $1.16 \mathrm{MeV}($ Mass equivalent $=0.00124 \mathrm{amu})$ of energy is absorbed. Mass on the reactant side is $18.00567$ amu and proton mass $=1.00782 \mathrm{amu}$. The atomic mass of ${ }^{17} \mathrm{O}$ will be

  • A
    $17.0044 \mathrm{amu}$
  • Correct answer
    B
    $16.9991 \mathrm{amu}$
  • C
    $17.0114 \mathrm{amu}$
  • D
    $16.9966 \mathrm{amu}$

Solution

${ }^{14} \mathrm{~N}_{7}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }^{17} \mathrm{O}_{8}+{ }_{1}^{1} \mathrm{H}$ mass defect $=0.00124 \mathrm{amu}$ $\left(\mathrm{m}_{\mathrm{o}^{17}}+\mathrm{m}_{\mathrm{H}^{1}}\right)-\left(\mathrm{m}_{\mathrm{N}^{44}}+\mathrm{m}_{\mathrm{He}^{4}}\right)=0.00124$ $\mathrm{~m}_{\mathrm{o}^{17}}+1.00782-[18.00567]=0.00124$ amu $\quad$ amu solving: $\mathrm{m}_{\mathrm{o}^{17}}=16.99909 \mathrm{amu}$ $\therefore 16.9991$ (approx)

Hence, the correct answer is option: 'B'.

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