JEE MAIN - Physics (2006 - No. 21)
If the binding energy per nucleon in $${}_3^7Li$$ and $${}_2^4He$$ nuclei are $$5.60$$ $$MeV$$ and $$7.06$$ $$MeV$$ respectively, then in the reaction
$$$p + {}_3^7Li \to 2\,{}_2^4He$$$
energy of proton must be
energy of proton must be
$$28.24$$ $$MeV$$
$$17.28$$ $$MeV$$
$$1.46$$ $$MeV$$
$$39.2$$ $$MeV$$
Explanation
Let $$E$$ be the energy of proton, then
$$E + 7 \times 5.6 = 2 \times \left[ {4 \times 7.06} \right]$$
$$ \Rightarrow E = 56.48 - 39.2 = 17.28MeV$$
$$E + 7 \times 5.6 = 2 \times \left[ {4 \times 7.06} \right]$$
$$ \Rightarrow E = 56.48 - 39.2 = 17.28MeV$$
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