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Is the fission of iron (""(26)Fe^(56)) ...

Is the fission of iron `(""_(26)Fe^(56))` into `(""_(13)Al^(28))` as given below possible?
`""_(26)Fe^(56)to ""_(13)Al^(28)+""_(13)Al^(28)+Q`
Given mass of `""_(26)Fe^(56)=55.934940 and ""_(13)Al^(28)=27.98191 U`

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AI Generated Solution

To determine if the fission of iron \( \left( {}_{26}^{56}\text{Fe} \right) \) into two aluminum nuclei \( \left( {}_{13}^{28}\text{Al} \right) \) is possible, we need to analyze the mass-energy balance of the reaction. ### Step-by-Step Solution: 1. **Identify the masses involved:** - Mass of iron \( \left( {}_{26}^{56}\text{Fe} \right) \): \( 55.934940 \, \text{u} \) - Mass of aluminum \( \left( {}_{13}^{28}\text{Al} \right) \): \( 27.98191 \, \text{u} \) ...
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Suppose, we think of fission of a ._26Fe^(56) nucleus into two equal fragments ._13Al^(28) . Is the fission energetically possible? Argue by working out Q of the process. Given m(._26Fe^(56))=55.93494u, m(._13Al^(28))=27.98191 u.

The nuclear reaction given below is of …… type ""_(13)^(27)Al + ""_(2)^(4)He to ""_(14)^(30)Si + ""_(1)^(1)H

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The neutron separation energy is defined to be the energy required to remove a neutron form nucleus. Obtain the neutron separartion energy of the nuclei ._(20)Ca^(41) and ._(13)Al^(27) form the following data : m(._20Ca^(40))=39.962591u and m(._(20)Ca^(41))=40.962278u m(._(13)Al^(26))=25.986895u and m(._(13)Al^(27))=26.981541u

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