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344 bytes added ,  19:28, January 21, 2013
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== Typical reaction ==
 
== Typical reaction ==
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When [[Uranium|Uranium-235]] <math>{}^{235}U</math> is bombarded with neutrons, a nucleus may absorb a neutron <math>{}_0^1n</math> and become very unstable. It splits into fragments, generally two new smaller nuclei and a few neutrons. A typical reaction would be:
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<center><math>{}^{235}_{92}U + {}_0^1n \qquad \rarr \qquad {}^{139}_{56}Ba + {}^{94}_{36}Kr + 3  {}_0^1n</math></center>
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{|class="wikitable" style="float: right;" border="1"
 
{|class="wikitable" style="float: right;" border="1"
 
!particle
 
!particle
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|align="right"|1.00866 amu
 
|align="right"|1.00866 amu
 
|}
 
|}
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When [[Uranium|Uranium-235]] <math>{}^{235}U</math> is bombarded with neutrons, a nucleus may absorb a neutron <math>{}_0^1n</math> and become very unstable. It splits into fragments, generally two new smaller nuclei and a few neutrons. A typical reaction would be:
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<center><math>{}^{235}_{92}U + {}_0^1n \qquad \rarr \qquad {}^{139}_{56}Ba + {}^{94}_{36}Kr + 3  {}_0^1n</math></center>
    
The reaction releases a great amount of [[binding energy]], as the binding energy of the <math>{}^{235}U</math> is much higher than the  binding energy of  <math>{}^{139}_{56}Ba</math> and <math>{}^{94}_{36}Kr</math> combined. This can be seen when we compare the masses of the elements involved on both sides of the equation: though the number of protons and neutrons on both sides is the same, the masses are different!
 
The reaction releases a great amount of [[binding energy]], as the binding energy of the <math>{}^{235}U</math> is much higher than the  binding energy of  <math>{}^{139}_{56}Ba</math> and <math>{}^{94}_{36}Kr</math> combined. This can be seen when we compare the masses of the elements involved on both sides of the equation: though the number of protons and neutrons on both sides is the same, the masses are different!
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Indeed, the particles of the left hand side have a weight of 236.0526[[amu]], those of the right hand side only 235.8692[[amu]]. The difference of 0.18341[[amu]] were turned into energy, according to Einstein's [[E=mc²]]:
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:<math>0.18341amu \cdot c^2 = 0.18341 \cdot 1.6605 \cdot 10^{−27} </math>kg \cdot 299,792,458 \frac{m^2}{s^2} = </math>
    
==References==
 
==References==
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