Answer :
The De-Broglie wavelength of an electron having kinetic energy 1200 keV is approximately [tex]1.12 \times 10^{-12}[/tex] meters, and the total energy released when 0.5 kg of U-235 undergoes fission is approximately [tex]4.27 \times 10^{13}[/tex] joules.
To solve these questions, we'll go step-by-step through both parts: calculating the De-Broglie wavelength of an electron and determining the energy released in a fission reaction of Uranium-235.
1. Calculating De-Broglie Wavelength:
The De-Broglie wavelength, λ, is given by the formula:
[tex]\lambda = \frac{h}{p}[/tex]
where:
- [tex]h[/tex] is Planck's constant ([tex]6.626 \times 10^{-34}[/tex] Js)
- [tex]p[/tex] is the momentum of the particle
First, we need to find the velocity of the electron. The kinetic energy (KE) is related to the velocity (v) by the equation:
[tex]KE = \frac{1}{2} mv^2[/tex]
where:
- [tex]m[/tex] is the mass of the electron ([tex]9.109383 \times 10^{-31}[/tex] kg)
- [tex]KE[/tex] is the given kinetic energy (1200 keV)
1 eV = J 1.60219 × 10⁻¹⁹
Therefore, 1200 keV = 1200 × 10³ × 1.60219 × 10⁻¹⁹
J = 1.9226 × 10⁻¹³ J
Now, solving for the velocity:
- 1/2mv² = 1.922×10⁻¹³
- v² = 2 × 1.9226 × 10⁻¹³ /9.109383 × 10⁻³¹
- v² = 4.220 × 10¹⁷
- v= 6.5 × 10⁸ m/s
The momentum [tex]p[/tex] can be found using:
p = mv
p = (9.109383 × 10⁻³¹) × (6.5 × 10⁸)
p = 5.92 × 10⁻²² kg m/s
Finally, the De-Broglie wavelength is:
[tex]\lambda = \frac{6.626 \times 10^{-34}}{ 5.92 \times 10^{-22}}[/tex]
[tex]\lambda = 1.12 \times 10^{-12} \text{ m}[/tex]
2. Energy Released in Fission of Uranium-235:
The energy released per fission of one Uranium-235 atom, [tex]Q[/tex], is given as 208 MeV.
- 1 MeV = [tex]1.60219 \times 10^{-13}[/tex] J
Therefore, 208 MeV = [tex]208 \times 1.60219 \times 10^{-13}[/tex] J = [tex]3.3346 \times 10^{-11}[/tex] J
We need to calculate the number of Uranium-235 atoms in 0.5 kg. The number of atoms can be found using the mass (m) and the molar mass (M) of Uranium-235:
- Molar mass of U-235 = 235 g/mol = [tex]0.235[/tex] kg/mol
- Number of atoms (N) = [tex]\frac{m}{M} \times N_A[/tex],
where [tex]N_A[/tex] is Avogadro's number 6.022 × 10²³ - Number of atoms in 0.5 kg of U-235:
- N = 0.5 / 0.235 × 6.022 × 10²³
- N = 1.28 × 10²⁴ atoms
Now, the total energy released (E) is:
- E= N × Q
- E = 1.28 × 10²⁴ × 3.3346 × 10⁻¹¹
- E = 4.27 × 10¹³ J
