# Difference between revisions of "Solutions 4"

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− | + | = Mass, Binding Energy and the Liquid Drop Model<br> = | |

− | + | ====== Return to [[Problem Solving Sets]] ====== | |

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+ | <br> '''1: '''<br> | ||

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+ | #Mass excess is given by <span class="texhtml">Δ</span><span class="texhtml" />M(Z,A) = M(Z,A)-A. The masses can therefore be found with M(Z,A) =<span class="texhtml"><span class="texhtml">Δ</span></span>M (Z,A ) + A by using 1 u = 931.5 MeV we get: <br>M(n) = 1.008665 u<br>M(<sup>1</sup>H) = 1.007825 u<br> M(<sup>4</sup>He) = 4.002603 u<br>M(<sup>56</sup>Fe) = 55.934938 u<br> M(<sup>142</sup>Ce) = 141.909197 u<br> M(<sup>238</sup>U) = 238.050788 u<br> | ||

#The most stable nuclei is <sup>56</sup>Fe. It has the highest binding energy compared to the number of nucleons. | #The most stable nuclei is <sup>56</sup>Fe. It has the highest binding energy compared to the number of nucleons. | ||

#1.00 kg <sup>2</sup>H is 496.5 mol by fusion 248,2 mol <sup>4</sup>He is created which is 993.6 g. The difference in mass is equal to the energy 1000 g-993.6 g = 6.36 g. The energy is then 5.7 <math>\cdot</math>10<sup>14</sup> J, 3.56 <math>\cdot</math> 10<sup>27</sup> MeV or 1.59 <math>\cdot</math>10<sup>8</sup> kWh. | #1.00 kg <sup>2</sup>H is 496.5 mol by fusion 248,2 mol <sup>4</sup>He is created which is 993.6 g. The difference in mass is equal to the energy 1000 g-993.6 g = 6.36 g. The energy is then 5.7 <math>\cdot</math>10<sup>14</sup> J, 3.56 <math>\cdot</math> 10<sup>27</sup> MeV or 1.59 <math>\cdot</math>10<sup>8</sup> kWh. | ||

− | #1.00 kg <sup>233</sup>U fission to <sup>92</sup>Rb, <sup>138</sup>Cs and three neutrons. The mass difference is 0.785 g. Which gives 7 | + | #1.00 kg <sup>233</sup>U fission to <sup>92</sup>Rb, <sup>138</sup>Cs and three neutrons. The mass difference is 0.785 g. Which gives 7.1 <math>\cdot</math>10<sup>13</sup> J, 4.40 <math>\cdot</math>10<sup>26</sup> MeV or 1.96 <math>\cdot</math>10<sup>7</sup> kWh. |

#The energy from the fission is distributed in different ways: most of it goes to the kinetic energy of for the fission products. Other parts goes to the kinetic energy of the neutrinos and the neutrons, and some of it goes to “prompt” gamma-rays and beta/gamma rays from the fission products. | #The energy from the fission is distributed in different ways: most of it goes to the kinetic energy of for the fission products. Other parts goes to the kinetic energy of the neutrinos and the neutrons, and some of it goes to “prompt” gamma-rays and beta/gamma rays from the fission products. | ||

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'''3:''' <br> | '''3:''' <br> | ||

− | <math>\frac{E_{nucleus}}{E_{electron}}=\frac{8111.493 \, keV}{5.14\cdot 10^{-3} \, keV}=1.6\cdot 10^{5}</math> times bigger binding energy for the nucleus.<br> | + | <math>\frac{E_{nucleus}}{E_{electron}}=\frac{8111.493 \, keV}{5.14\cdot 10^{-3} \, keV}=1.6\cdot 10^{5}</math> |

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+ | 1.6<math>\cdot</math>10<sup>5 </sup>times bigger binding energy for the nucleus.<br> | ||

<br> | <br> | ||

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'''4:''' <br> | '''4:''' <br> | ||

− | # | + | #m<sub>n</sub> = 1.67 <math>\cdot</math>10<sup>-27</sup>kg |

− | # | + | #m<sub>e</sub> = 9.11 <math>\cdot</math>10<sup>-31</sup>kg<br> |

− | # | + | #m<sub>"u"</sub> = 1.66 <math>\cdot</math>10<sup>-27</sup>kg<br> |

<br> | <br> | ||

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<br> | <br> | ||

− | '''6:''' The energy | + | '''6:''' |

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+ | The energy | ||

<span class="texhtml">M(<sup>235</sup>U) - (M(<sup>131</sup>Xe + M(<sup>101</sup>Ru) + 3M(n)) = </span> | <span class="texhtml">M(<sup>235</sup>U) - (M(<sup>131</sup>Xe + M(<sup>101</sup>Ru) + 3M(n)) = </span> | ||

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<span class="texhtml"><span class="texhtml"><span class="texhtml"> = ΔM(<sup>235</sup>U) - ΔM(<sup>131</sup>Xe) - ΔM(<sup>101</sup>Ru) - 3ΔM(n)</span></span></span> | <span class="texhtml"><span class="texhtml"><span class="texhtml"> = ΔM(<sup>235</sup>U) - ΔM(<sup>131</sup>Xe) - ΔM(<sup>101</sup>Ru) - 3ΔM(n)</span></span></span> | ||

− | <span class="texhtml"><span class="texhtml"><span class="texhtml">=40.916 + 88.421 + 87.952 - 3</span></span></span><math>\cdot</math> 8.071 = 193.08 | + | <span class="texhtml"><span class="texhtml"><span class="texhtml">=40.916 + 88.421 + 87.952 - 3</span></span></span><math>\cdot</math> 8.071 = 193.08 MeV<br> |

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+ | <br> | ||

'''7:'''<br> | '''7:'''<br> | ||

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1 g <sup>235</sup>U=4.25<math>\cdot</math>10<sup>-3</sup> mol = 2.56<math>\cdot</math>10<sup>21</sup> atoms. <br> | 1 g <sup>235</sup>U=4.25<math>\cdot</math>10<sup>-3</sup> mol = 2.56<math>\cdot</math>10<sup>21</sup> atoms. <br> | ||

− | If all of the atoms fission and assuming 100% efficiency this will give 5.12<math>\cdot</math>10<sup>29</sup> eV = 8.22<math>\cdot</math>10<sup>7</sup>kJ. Which will last for 135190 km.<br> | + | If all of the atoms fission and assuming 100% efficiency this will give 5.12 <math>\cdot</math>10<sup>29</sup> eV = 8.22 <math>\cdot</math>10<sup>7</sup>kJ. Which will last for 135190 km.<br> |

<br> | <br> | ||

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'''8:'''<br> | '''8:'''<br> | ||

− | <span class="texhtml"> | + | <span class="texhtml">ΔG = ΔG<sub>product</sub> - ΔG<sub>reactant</sub>= - 237 kJ/mol</span>. Which means that for each H2O molecule it is generated 237kJ/NA = 2.456<math>\cdot</math>10<sup>-6</sup>MeV. Comparing to 0.0303 u = 23.2 MeV released by fusion to helium. We see that the nuclear reactions functions on a scale 10<sup>6</sup> times bigger than chemical. |

<br> | <br> | ||

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'''9:''' | '''9:''' | ||

− | + | Fusion of <sup>2</sup>H gives 5.75 <math>\cdot</math>10<sup>11</sup> J/gram approximately ten times as much energy per fusion than uranium which gives 8.2 <math>\cdot</math>10<sup>10</sup> J/gram.<br> | |

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+ | <br> | ||

'''10:''' | '''10:''' | ||

− | In this problem it is | + | In this problem it is useful to keep in mind the binding energy equation:<br> |

<math>B_{tot}(A,Z)=a_{v}A-a_{s}A^{2/3}-a_{c}\frac{Z^{2}}{A^{1/3}}-a_{a}\frac{A-2Z}{A}\pm\delta</math> | <math>B_{tot}(A,Z)=a_{v}A-a_{s}A^{2/3}-a_{c}\frac{Z^{2}}{A^{1/3}}-a_{a}\frac{A-2Z}{A}\pm\delta</math> | ||

− | Most isobars (A= constant, Z varies) will have binding energy when plotted as a function of Z creating a parable. The last part of the equation in the formula for binding energy delta expresses the energy gained when the proton and/or the neutron are in a couple. A nucleus with a even amount of both will have a | + | Most isobars (A= constant, Z varies) will have binding energy when plotted as a function of Z creating a parable. The last part of the equation in the formula for binding energy delta expresses the energy gained when the proton and/or the neutron are in a couple. A nucleus with a even amount of both will have a positive <span class="texhtml">δ</span>, higher binding energy, a unpaired nucleus either proton or neutron will have a <span class="texhtml">δ</span> of zero, a nucleus with a unpaired neutron and proton will have a negative <span class="texhtml">δ</span>. Therefore for odd-nuclei the binding energy is independent of which nucleon is unpaired. The binding energy as a function of z will have a minimum value for a certain stable nucleus. For even nuclei with a given A value they will have a even-even, or a odd-odd configuration when Z varies. This will create two curves where the odd-odd nuclei will have a lower binding energy than the even-even nuclei. Odd-odd nuclei will therefore have more possibilities when decaying to a level with higher binding energy, either with a β<sup>-</sup> decay or a β<sup>+</sup> decay. Both of these mechanisms will lead to a more stable nucleus. |

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+ | <br> | ||

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+ | '''11:''' | ||

− | + | answered above<br> | |

− | + | [[Category:Solved_Problem]] [[Category:Bachelor]] |

## Latest revision as of 10:02, 9 July 2012

# Mass, Binding Energy and the Liquid Drop Model

###### Return to Problem Solving Sets

**1: **

- Mass excess is given by Δ<span class="texhtml" />M(Z,A) = M(Z,A)-A. The masses can therefore be found with M(Z,A) =ΔM (Z,A ) + A by using 1 u = 931.5 MeV we get:

M(n) = 1.008665 u

M(^{1}H) = 1.007825 u

M(^{4}He) = 4.002603 u

M(^{56}Fe) = 55.934938 u

M(^{142}Ce) = 141.909197 u

M(^{238}U) = 238.050788 u

- The most stable nuclei is
^{56}Fe. It has the highest binding energy compared to the number of nucleons. - 1.00 kg
^{2}H is 496.5 mol by fusion 248,2 mol^{4}He is created which is 993.6 g. The difference in mass is equal to the energy 1000 g-993.6 g = 6.36 g. The energy is then 5.7 10^{14}J, 3.56 10^{27}MeV or 1.59 10^{8}kWh. - 1.00 kg
^{233}U fission to^{92}Rb,^{138}Cs and three neutrons. The mass difference is 0.785 g. Which gives 7.1 10^{13}J, 4.40 10^{26}MeV or 1.96 10^{7}kWh. - The energy from the fission is distributed in different ways: most of it goes to the kinetic energy of for the fission products. Other parts goes to the kinetic energy of the neutrinos and the neutrons, and some of it goes to “prompt” gamma-rays and beta/gamma rays from the fission products.

**2: **

binding energy B per nucleon:

It is 0.00868 u per nucleon

and 8.26 MeV per nucleon

**3:**

1.6^{5 }times bigger binding energy for the nucleus.

**4:**

- m
_{n}= 1.67 10^{-27}kg - m
_{e}= 9.11 10^{-31}kg

- m
_{"u"}= 1.66 10^{-27}kg

**5:**

- 8.55 MeV/nucleon.

- 8.79 MeV/nucleon.

- 7.86 MeV/nucleon.

**6:**

The energy

M(^{235}U) - (M(^{131}Xe + M(^{101}Ru) + 3M(n)) =

(ΔM(^{235}U) + 235) - (Δ(^{131}Xe) + 131) - (ΔM(^{101}Ru) + 101) - 3(ΔM(m) + 1)

= ΔM(^{235}U) - ΔM(^{131}Xe) - ΔM(^{101}Ru) - 3ΔM(n)

=40.916 + 88.421 + 87.952 - 3 8.071 = 193.08 MeV

**7:**

Energy usage per 10 km:

1 L = 700 g. M_{m}(C_{8}H_{18}) = 114 g/mol, this means that 700 g is 6.14 mol.

6.14 mol

1 g ^{235}U=4.25 10^{-3} mol = 2.56 10^{21} atoms.

If all of the atoms fission and assuming 100% efficiency this will give 5.12 ^{29} eV = 8.22 10^{7}kJ. Which will last for 135190 km.

**8:**

ΔG = ΔG_{product} - ΔG_{reactant}= - 237 kJ/mol. Which means that for each H2O molecule it is generated 237kJ/NA = 2.456 10^{-6}MeV. Comparing to 0.0303 u = 23.2 MeV released by fusion to helium. We see that the nuclear reactions functions on a scale 10^{6} times bigger than chemical.

**9:**

Fusion of ^{2}H gives 5.75 10^{11} J/gram approximately ten times as much energy per fusion than uranium which gives 8.2 10^{10} J/gram.

**10:**

In this problem it is useful to keep in mind the binding energy equation:

Most isobars (A= constant, Z varies) will have binding energy when plotted as a function of Z creating a parable. The last part of the equation in the formula for binding energy delta expresses the energy gained when the proton and/or the neutron are in a couple. A nucleus with a even amount of both will have a positive δ, higher binding energy, a unpaired nucleus either proton or neutron will have a δ of zero, a nucleus with a unpaired neutron and proton will have a negative δ. Therefore for odd-nuclei the binding energy is independent of which nucleon is unpaired. The binding energy as a function of z will have a minimum value for a certain stable nucleus. For even nuclei with a given A value they will have a even-even, or a odd-odd configuration when Z varies. This will create two curves where the odd-odd nuclei will have a lower binding energy than the even-even nuclei. Odd-odd nuclei will therefore have more possibilities when decaying to a level with higher binding energy, either with a β^{-} decay or a β^{+} decay. Both of these mechanisms will lead to a more stable nucleus.

**11:**

answered above