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.

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

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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1. According to Einstein holds E=m·c² or ΔE=Δm·c²
2. c is the speed of light, 300'000 km/s
3. Transform the Einstein equation according to m or Δm
4. Δm = ΔE/c²
5. Δm = 3.82·10²⁶[kg·m²/s²]/(300'000'000 m/s)² = 4.24·10⁹ kg

The loss of mass per [translation unavailable] (!) is thus 4.24·10⁹ kg or approx. 424 millions of which:.

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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Our sun has a power of 3.82·10²⁶ W. So what is the mass loss per second?
Note: [W] = [J/s] = [kg·m²/s²]

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See above all the solution in the attached pdf.
Indications: at the exercise a) is the measured bq-value Even before the first half-life. At exercise b) the measured Bq value is between the first and second half-life.

Worked solution [here] as [translation unavailable]

In a fresh cotton fabric, 0.27 Bq is measured.
a) The Shroud of Turin measures 0.25 Bq. How old is the Shroud?
(b) In another tissue, 0.1 Bq is measured.
It is assumed that a linear decay occurs between two half-lives.

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Nothing has been done yet.

In a living organism, you measure about 16 decay per second per gram of carbon. In an old piece of wood that contains 5 g of carbon, you measure 44 decay per second. How old is the piece of wood?
Assumption: linear decay between two half-lives, T1_2(C) is 5730 years.

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