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Show that ""(92)^(238)U can not spontane...

Show that `""_(92)^(238)U` can not spontaneously emit a proton. Given:
`""_(92)^(238)U` = 238.05079u, `""_(91)^(237)Pa` = 237.05121u `""_(1)^(1)H` = 1.00783u

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To show that \( _{92}^{238}U \) cannot spontaneously emit a proton, we will analyze the reaction energetically. The reaction can be represented as follows: \[ _{92}^{238}U \rightarrow _{91}^{237}Pa + _{1}^{1}H + Q \] Where: - \( _{92}^{238}U \) is Uranium-238 ...
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What is the difference between ""_(92)"U"^(235) and ""_(92)"U"^(238) atoms ?

Calculate the energy released in MeV in the following nuclear reaction : ._(92)^(238)Urarr._(90)^(234)Th+._(2)^(4)He+Q ["Mass of "._(92)^(238)U=238.05079 u Mass of ._(90)^(238)Th=234.043630 u Massof ._(2)^(4)He=4.002600 u 1u = 931.5 MeV//c^(2)]

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