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Lösning 5.5:3 - Versionshistorik
2026-04-08T06:08:25Z
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Louwah: Ny sida: a) <math>\mathrm{Q = (M_{2H} + M_{3H} –\, M_{He} + m_n)c^2}</math> <math>\mathrm{(M_{2H})c^2 = (M_H + m_n)c^2 – 2\cdot 1{,}16\, MeV}</math> <math>\mathrm{(M_{3H})c^2 = (M_H + 2m_n)c^2...
2017-12-13T14:40:03Z
<p>Ny sida: a) <math>\mathrm{Q = (M_{2H} + M_{3H} –\, M_{He} + m_n)c^2}</math> <math>\mathrm{(M_{2H})c^2 = (M_H + m_n)c^2 – 2\cdot 1{,}16\, MeV}</math> <math>\mathrm{(M_{3H})c^2 = (M_H + 2m_n)c^2...</p>
<p><b>Ny sida</b></p><div>a) <math>\mathrm{Q = (M_{2H} + M_{3H} –\, M_{He} + m_n)c^2}</math><br />
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<math>\mathrm{(M_{2H})c^2 = (M_H + m_n)c^2 – 2\cdot 1{,}16\, MeV}</math><br />
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<math>\mathrm{(M_{3H})c^2 = (M_H + 2m_n)c^2 – 3\cdot 2{,}83\, MeV}</math><br />
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<math>\mathrm{(M_{He})c^2 = (2M_H + 2m_n)c^2 – 4\cdot 7{,}07\, MeV}</math><br />
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vilket ger <math>\textrm{Q} = -2\cdot 1{,}16 -3\cdot 2{,}83 - (-7\cdot 7{,}07) = 17{,}47 \textrm{ MeV}</math><br />
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b) <math>\mathrm{U = e^2/4\pi \varepsilon_0(R_{2H} + R_{3H}) = (R = R_0\, A^{1/3},\, R_0 = 1,4\, fm) = 380\, keV}</math></div>
Louwah