Mass loss 2 (Radio-51) It will be a grams 23592U in full split. (a) Calculate the energy released per gram. For each split core, 200 MeV energy is released. Note: 1ev = 1.602·10-19 J (b) How high could you lift a car (m = 1000 kg) with this energy?

  Exercise: Radio-51    

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It will be a grams 23592U in full split.
(a) Calculate the energy released per gram. For each split core, 200 MeV energy is released.

    Note: 1ev = 1.602·10^-19 J

(b) How high could you lift a car (m = 1000 kg) with this energy?

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Solution:

1. n = m/M
2. n( 23592U) = m( 23592U)/M( 23592U) = 1 g/235 g/mol = 0.0043 mol
3. The number of atoms in the atomic nucleus is 6.022·10²³·0.0043 = 2.563·10²¹
4. for each split 200 MeV or 200·10⁶·1.602·10^-19 J = 3.204·10^-11 J

(a) Total is 2.56·10²¹ · 3.204·10^-11 J = 8.21·10¹⁰ J
b) with the energy can what made are, z.B. [translation unavailable] lifting: E=m·g·h

    h =E/(m·g) = 8.21·10¹⁰/(1000·10) = 8210000 m = 8210 km

    To the comparison height mount Everest approx. 8 km on sea, thus 1000· car lifting

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