Mass loss 3 (Radio-52) It will be a grams 23592U in full split. Per grams will be approx. 8.21·1010 J energy free, see in addition also a previously exercise. The switzerland required in the the last 5 years average round 810'000 terahjoule energy per year. Suppose you only want to gain this energy by splitting 23592U. How big would an appropriate uranium bullet be? Further data are needed for the calculation: 1 terajoule = 1012 joules, total incense: 8.1·1017 J Other sources say 2.04·1017 J Density uranium ores 19.1 g/cm3

  Exercise: Radio-52    

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It will be a grams 23592U in full split. Per grams will be approx. 8.21·10¹⁰ J energy free, see in addition also a previously exercise. The switzerland required in the the last 5 years average round 810'000 terahjoule energy per year. Suppose you only want to gain this energy by splitting 23592U. How big would an appropriate uranium bullet be?
Further data are needed for the calculation:

1. 1 terajoule = 10¹² joules, total incense: 8.1·10¹⁷ J
2. Other sources say 2.04·10¹⁷ J
3. Density uranium ores 19.1 g/cm³

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Solution:

1. Terahjoule = 10 ¹²J
       Annual consumption 810'000 terajoules = 0.81·10⁶ TJ = 0.81·¹⁸ J = 8.1·10¹⁷ J
2. for each gram 8.21·10¹⁰ J
       weight required: 8.1·10¹⁷ J / 8.21·10¹⁰ J = 9.87·10⁶ grams
3. density = D = m/V
       V = m/D = 9.87·10⁶ g / 19.1 g/cm³ = 516'754 cm³
4. V(Kugel) = 4/3·pi·r³
       R = 3 roots from (3·V/4·pi) = (3·516754/4·pi)^0.33333 = 49.7 cm
       So the diameter of the ball is about 1 meter.

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