Louwah: Ny sida: a) $\mathrm{Q = (M_{2H} + M_{3H} –\, M_{He} + m_n)c^2}$ $\mathrm{(M_{2H})c^2 = (M_H + m_n)c^2 – 2\cdot 1{,}16\, MeV}$ $\mathrm{(M_{3H})c^2 = (M_H + 2m_n)c^2...$

Louwah — Wed, 13 Dec 2017 14:40:03 GMT

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...

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 – 3\cdot 2{,}83\, MeV}</math>

<math>\mathrm{(M_{He})c^2 = (2M_H + 2m_n)c^2 – 4\cdot 7{,}07\, MeV}</math>

vilket ger <math>\textrm{Q} = -2\cdot 1{,}16 -3\cdot 2{,}83 - (-7\cdot 7{,}07) = 17{,}47 \textrm{ MeV}</math>

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>

Lösning 5.5:3 - Versionshistorik

Louwah: Ny sida: a) \mathrm{Q = (M_{2H} + M_{3H} –\, M_{He} + m_n)c^2} \mathrm{(M_{2H})c^2 = (M_H + m_n)c^2 – 2\cdot 1{,}16\, MeV} \mathrm{(M_{3H})c^2 = (M_H + 2m_n)c^2...

Louwah: Ny sida: a) $\mathrm{Q = (M_{2H} + M_{3H} –\, M_{He} + m_n)c^2}$ $\mathrm{(M_{2H})c^2 = (M_H + m_n)c^2 – 2\cdot 1{,}16\, MeV}$ $\mathrm{(M_{3H})c^2 = (M_H + 2m_n)c^2...$