Respuesta :
Answer:
The sum of the kinetic energies of the alpha particle and the new nucleus = (1.359098 × 10⁻¹²) J
Explanation:
We will use the conservation of energy theorem for extremely small particles,
Total energy before split = total energy after split
That is,
Total energy of the original nucleus = (total energy of the new nucleus) + (total energy of the alpha particle)
Total energy of these subatomic particles is given as equal to (rest energy) + (kinetic energy)
Rest energy = mc² (Einstein)
Let Kinetic energy be k
Kinetic energy of original nucleus = k₀ = 0 J
Kinetic energy of new nucleus = kₙ
Kinetic energy of alpha particle = kₐ
Mass of original nucleus = m₀ = (3.969554 × 10⁻²⁵) kg
Mass of new nucleus = mₙ = (3.902996 × 10⁻²⁵) kg
Mass of alpha particle = mₐ = (6.640678 × 10⁻²⁷) kg
Speed of light = (2.9979246 × 10⁸) m/s
Total energy of the original nucleus = m₀c² (kinetic energy = 0, since it was originally at rest)
Total energy of new nucleus = (mₙc²) + kₙ
Total energy of the alpha particle = (mₐc²) + kₐ
(m₀c²) = (mₙc²) + kₙ + (mₐc²) + kₐ
kₙ + kₐ = (m₀c²) - [(mₙc²) + (mₐc²)
(kₙ + kₐ) = c² (m₀ - mₙ - mₐ)
(kₙ + kₐ) = (2.9979246 × 10⁸)² [(3.969554 × 10⁻²⁵) - (3.902996 × 10⁻²⁵) - (6.640678 × 10⁻²⁷)]
(kₙ + kₐ) = (8.98755191 × 10¹⁶)(1.5122 × 10⁻²⁹) = (1.35909760 × 10⁻¹²) J
