If the binding energy per nucleon in -7-textrm-3-Li and He nuclei are 5-60 MeV and 7-06 MeV respectively- then in the reaction p-3-textrm-7-Lirightarrow -2-textrm-4- He- binding energy is39-2 MeV17-28 MeV28-24 MeV1-46 MeV

Question

Toppr Admin · Accepted Answer

Since- p-3Li7-x2192-22He4-xA0-Thus- proton energy - BE of He atom -xA0-x2212- BE of Li of atom-or Ep-2-xD7-4-xD7-7-06-x2212-7-xD7-5-60-17-28MeV -xA0-where 4 is the no of nucleon of He and 7 is the no of nucleon of Li-xA0-

If the binding energy per nucleon in $_{7} 3 L i$ and He nuclei are $5.60 M e V$ and $7.06 M e V$ respectively, then in the reaction $p +_{3} 7 L i \to_{2} 4$ He, binding energy is
39.2 MeV
17.28 MeV
28.24 MeV
1.46 MeV

Since, $p +_{3} L i^{7} \to 2_{2} H e^{4}$
Thus, proton energy = BE of He atom $-$ BE of Li of atom.
or $E_{p} = (2 \times 4 \times 7.06) - (7 \times 5.60) = 17.28 M e V$ where 4 is the no of nucleon of He and 7 is the no of nucleon of Li.

If the binding energy per nucleon in 73Li and He nuclei are 5.60MeV and 7.06MeV respectively, then in the reaction p+37Li→24 He, binding energy is39.2 MeV17.28 MeV28.24 MeV1.46 MeV

Since, p+3Li7→22He4 Thus, proton energy = BE of He atom − BE of Li of atom.or Ep=(2×4×7.06)−(7×5.60)=17.28MeV where 4 is the no of nucleon of He and 7 is the no of nucleon of Li.

If the binding energy per nucleon in $_{7} 3 L i$ and He nuclei are $5.60 M e V$ and $7.06 M e V$ respectively, then in the reaction $p +_{3} 7 L i \to_{2} 4$ He, binding energy is
39.2 MeV
17.28 MeV
28.24 MeV
1.46 MeV

Since, $p +_{3} L i^{7} \to 2_{2} H e^{4}$
Thus, proton energy = BE of He atom $-$ BE of Li of atom.
or $E_{p} = (2 \times 4 \times 7.06) - (7 \times 5.60) = 17.28 M e V$ where 4 is the no of nucleon of He and 7 is the no of nucleon of Li.