Question-154554

Question Number 154554 by peter frank last updated on 19/Sep/21

Commented by peter frank last updated on 19/Sep/21

$$\:\:_{\mathrm{2}} ^{\mathrm{3}} \mathrm{He}=\mathrm{3}.\mathrm{01664U} \\ $$

Answered by peter frank last updated on 19/Sep/21

Calculate the binding energy per nuclei for the _2^4 He and _2^3 He

$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{binding}\:\mathrm{energy} \\ $$$$\mathrm{per}\:\mathrm{nuclei}\:\mathrm{for}\:\mathrm{the}\:\:_{\mathrm{2}} ^{\mathrm{4}} \mathrm{He}\:\mathrm{and}\:_{\mathrm{2}} ^{\mathrm{3}} \mathrm{He} \\ $$

Answered by peter frank last updated on 20/Sep/21

_1^1 H=1.007834U _2^3 He=3.01664U _2^4 He=4.00387U _0^1 η=1.008674U _2^4 He→2 _1^1 H+2_0 ^1 η _2^4 He(mass)=m_1 2 _1^1 H+2_0 ^1 η(mass)=m_2 △m=m_1 −m_2 =0.029134U from 1U=931Mev 0.029134U=? Binding energ=931×0.029134U=27.12Mev Binding energy per nucleon =((27.12)/4)=6.78Mev/nucleon _2^3 He_(m_1 ) →2 _1^1 H+2_0 ^1 η_(m_2 ) △m=m_1 −m_2 =7.69×10^(−3) U 1U=931Mev 7.69×10^(−3) U=? B.E=7.15939Mev B.E/Nucleon=((7.15939Mev)/3)=2.386Mev/nucleon

$$\:\:_{\mathrm{1}} ^{\mathrm{1}} \mathrm{H}=\mathrm{1}.\mathrm{007834U} \\ $$$$\:\:_{\mathrm{2}} ^{\mathrm{3}} \mathrm{He}=\mathrm{3}.\mathrm{01664U} \\ $$$$\:\:_{\mathrm{2}} ^{\mathrm{4}} \mathrm{He}=\mathrm{4}.\mathrm{00387U} \\ $$$$\:_{\mathrm{0}} ^{\mathrm{1}} \eta=\mathrm{1}.\mathrm{008674U} \\ $$$$\:_{\mathrm{2}} ^{\mathrm{4}} \mathrm{He}\rightarrow\mathrm{2}\:_{\mathrm{1}} ^{\mathrm{1}} \mathrm{H}+\mathrm{2}_{\mathrm{0}} ^{\mathrm{1}} \eta \\ $$$$\:_{\mathrm{2}} ^{\mathrm{4}} \mathrm{He}\left(\mathrm{mass}\right)=\mathrm{m}_{\mathrm{1}} \\ $$$$\mathrm{2}\:_{\mathrm{1}} ^{\mathrm{1}} \:\mathrm{H}+\mathrm{2}_{\mathrm{0}} ^{\mathrm{1}} \eta\left(\mathrm{mass}\right)=\mathrm{m}_{\mathrm{2}} \\ $$$$\bigtriangleup\mathrm{m}=\mathrm{m}_{\mathrm{1}} −\mathrm{m}_{\mathrm{2}} =\mathrm{0}.\mathrm{029134U} \\ $$$$\mathrm{from} \\ $$$$\mathrm{1U}=\mathrm{931Mev} \\ $$$$\mathrm{0}.\mathrm{029134U}=? \\ $$$$\mathrm{Binding}\:\mathrm{energ}=\mathrm{931}×\mathrm{0}.\mathrm{029134U}=\mathrm{27}.\mathrm{12Mev} \\ $$$$\mathrm{Binding}\:\mathrm{energy}\:\mathrm{per}\:\mathrm{nucleon} \\ $$$$=\frac{\mathrm{27}.\mathrm{12}}{\mathrm{4}}=\mathrm{6}.\mathrm{78Mev}/\mathrm{nucleon} \\ $$$$\underset{\mathrm{m}_{\mathrm{1}} } {\underbrace{\:_{\mathrm{2}} ^{\mathrm{3}} \mathrm{He}}}\rightarrow\underset{\mathrm{m}_{\mathrm{2}} } {\underbrace{\mathrm{2}\:_{\mathrm{1}} ^{\mathrm{1}} \mathrm{H}+\mathrm{2}_{\mathrm{0}} ^{\mathrm{1}} \eta}} \\ $$$$\bigtriangleup\mathrm{m}=\mathrm{m}_{\mathrm{1}} −\mathrm{m}_{\mathrm{2}} =\mathrm{7}.\mathrm{69}×\mathrm{10}^{−\mathrm{3}} \:\mathrm{U} \\ $$$$\mathrm{1U}=\mathrm{931Mev} \\ $$$$\mathrm{7}.\mathrm{69}×\mathrm{10}^{−\mathrm{3}} \:\mathrm{U}=? \\ $$$$\mathrm{B}.\mathrm{E}=\mathrm{7}.\mathrm{15939Mev} \\ $$$$\mathrm{B}.\mathrm{E}/\mathrm{Nucleon}=\frac{\mathrm{7}.\mathrm{15939Mev}}{\mathrm{3}}=\mathrm{2}.\mathrm{386Mev}/\mathrm{nucleon} \\ $$

Commented by Tawa11 last updated on 21/Sep/21

$$\mathrm{nice}\:\mathrm{sir} \\ $$

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