| Consider following nuclear reactions and their reaction energies Q. |
| \[^{14}N+{}^{2}H\xrightarrow{{}}{}^{15}N+{}^{1}H+8.53Me\,V\] |
| \[^{15}N+{}^{2}H\xrightarrow{{}}{}^{13}C+{}^{4}He+7.58Me\,V\] |
| \[^{13}C+{}^{2}H\xrightarrow{{}}{}^{11}B+{}^{4}He+5.16Me\,V\] |
| It is given that, masses of involved nuclii are |
| \[_{2}^{4}He=4.0026u,\] \[_{1}^{2}H=2.0140u,\] |
| \[_{1}^{1}H=1.0078\,u,\] \[_{0}^{1}n=1.0087u\] |
| \[1u=931\frac{MeV}{{{c}^{2}}}\] |
| Using above data, Q value of reaction |
| \[{}^{11}B+{}^{4}He\xrightarrow{{}}{}^{14}N+_{0}^{1}n\] is |
A) 0.5 MeV
B) 0.005 MeV
C) 0.05 MeV
D) 0.5 eV
Correct Answer: C
Solution :
| First reaction is \[{}^{14}N+{}^{2}H\xrightarrow{{}}{}^{15}N+{}^{1}H+8.53MeV\] | |
| We can rearrange above equation as, \[\Rightarrow \]\[{}^{14}N-{}^{15}N\xrightarrow{{}}p-d+8.53\] | |
| where, \[p={}^{1}H,d={}^{2}H\] | |
| So, rewriting given reactions, we have | |
| \[{}^{14}N-{}^{15}N=p-d+8.53\] | ?(i) |
| \[{}^{15}N-{}^{13}C=\alpha -d+7.58\] | ?(ii) |
| \[{}^{13}C-{}^{11}B=\alpha -d+5.16\] | ?(iii) |
| Adding above equations, we have |
| \[{}^{14}N-{}^{11}B=p+2\alpha -3d+21.27\] |
| \[\Rightarrow \]\[{}^{11}B-{}^{14}N+\alpha -n\]\[=3d-\alpha -p-n-21.27\] |
| Now, \[3d-\alpha -p-n\]\[=3\times 2.0140-4.0026-1.0078-1.0087\]\[=0.0229u\] |
| So, Q value for reaction is \[{}^{11}B{{(\alpha ,n)}^{14}}N=0.229\times 931-21.27\]\[=0.05MeV\] |
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