Question

The following deuterium reactions and corresponding reaction energies are found to occur
${}^{14}N(d,p{)}^{15}N,Q=8.53MeV$
${}^{15}N(d,\alpha {)}^{13}C,Q=7.58MeV$
${}^{13}C(d,\alpha {)}^{11}B,Q=5.16MeV$
The rotation ${}^{14}N(d,p{)}^{15}N$ represents the reaction ${}^{14}N+d{\to }^{15}N+p$
${}_2^{4}He=4.0026amu{,}_1^{2}He=2.014amu{,}_1^{1}H=1.0078amu,n=1.0087amu(1amu=931MeV)$
The Q values of the reaction ${}^{11}B(\alpha ,n{)}^{14}N$ is

• A. $0.5eV$
• B. $0.5MeV$
• C. $0.05MeV$
• D. $0.05eV$

Answer 1

Answer: (C)

$0.05MeV$

The following reactions are as follows-

${}^{14}N(d,p{)}^{15}N{:}^{14}N+d{\to }^{15}N+p{Q}_1=8.53MeV$ ..............(i)

${}^{15}N(d,\alpha {)}^{13}C{:}^{15}N+d{\to }^{13}C+\alpha Q_2=7.58MeV$ ..............(ii)

${}^{13}C(d,\alpha {)}^{11}B{:}^{13}C+d{\to }^{11}B+\alpha Q_2=5.16MeV$ ...........(iii)

Adding (i), (ii) $\left(iii\right),$\implies$ ${}^{14}N+3d{\to }^{11}B+2\alpha +p$ ...........(a)

$Q$ value for this reaction $Q_a=Q_1+Q_2+Q_3=21.27MeV$

Now for reaction, $3d\to p+\alpha +n$ ...........(b)

$\therefore Q_b=[3\times 2.014-1.0078+4.0026-1.0087]\times 931MeV=21.32MeV$

Now $\left(b\right)-\left(a\right)⟹$ ${}^{11}B+\alpha {\to }^{14}N+n$

$\therefore$ Q value for this reaction, $Q=Q_b-Q_a=21.32-21.27=0.05MeV$