Shortcut Methods

(A) Fission

Shortcut Methods and Tricks:

  • To calculate the energy released by the fission of 1 kg of uranium-235, use the following formula:

$$E = (1\text{ kg})(10^3 \text{ g})(1.661 \times 10^{-27} \text{ kg/g})(1014 \text{ kJ/kg}) = 1.67 \times 10^{13} \text{ kJ}$$

  • To calculate the critical mass of U235 for an uncontrolled chain reaction, use the following formula:

$$M_c = \frac{4\pi R^3 \rho}{3 k_{eff}^2}$$

where:

  • M_c is the critical mass (in kg)
  • R is the radius of the sphere of U235 (in m)
  • ρ is the density of U235 (in kg/m3)
  • k_eff is the effective multiplication factor

Example:

Calculate the critical mass of U235 for an uncontrolled chain reaction in a sphere with a radius of 10 cm. Assume that the density of U235 is 19.1 g/cm3 and the effective multiplication factor is 1.2.

$$M_c = \frac{4\pi (0.1 \text{ m})^3 (19.1 \text{ g/cm}^3)(1000 \text{ kg/g})}{3 (1.2)^2} = 49.5 \text{ kg}$$

Therefore, the critical mass of U235 for an uncontrolled chain reaction in a sphere with a radius of 10 cm is 49.5 kg.


#(B) Fusion Shortcut Methods and Tricks:

  • To calculate the energy released by the fusion of two deuterium atoms, use the following formula:

$$E = 3.2\text{ MeV} = 3.2 \times 10^6 \text{ eV} = 5.12 \times 10^{-13} \text{ J}$$

  • To calculate the energy released by the fusion of two tritium atoms, use the following formula:

$$E = 17.6\text{ MeV} = 17.6 \times 10^6 \text{ eV} = 2.82 \times 10^{-12} \text{ J}$$

Example:

Calculate the total energy released by the fusion of 1020 deuterium atoms and 1020 tritium atoms.

$$E_{total} = (10^{20} \text{ atoms})(3.2 \times 10^{-13} \text{ J/atom}) + (10^{20} \text{ atoms})(2.82 \times 10^{-12} \text{ J/atom})$$

$$E_{total} = 3.14 \times 10^7 \text{ J} + 2.82 \times 10^8 \text{ J} = 3.13 \times 10^8 \text{ J}$$

Therefore, the total energy released by the fusion of 1020 deuterium atoms and 1020 tritium atoms is 3.13 × 108 J.


#(C) Radioactivity Shortcut Methods and Tricks:

  • To calculate the number of radioactive atoms at time t, use the following formula:

$$N_t = N_0 e^{-\lambda t}$$

where:

  • N_t is the number of radioactive atoms at time t
  • N_0 is the original number of radioactive atoms
  • λ is the decay constant

Example:

A sample of radioactive material contains 1000 atoms. If the decay constant is 0.01 s^-1, how many radioactive atoms will remain after 10 seconds?

$$N_t = 1000 e^{-(0.01 \text{ s}^{-1})(10 \text{ s})} = 1000 e^{-0.1} = 905$$

Therefore, 905 radioactive atoms will remain after 10 seconds.


Note: Please note that these shortcut methods and tricks may not be applicable in all cases, so it is important to have a good understanding of the underlying principles and equations involved in these topics.



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