In a location in outer space far from all other objects, a nucleus whose mass is 3.969554 * 10-25 kg and that is initially at rest undergoes spontaneous alpha decay. The original nucleus disappears, and two new particles appear: a He-4 nucleus of mass 6.640678 * 10-27 kg (an alpha particle consisting of two protons and two neutrons) and a new nucleus of mass 3.902996 * 10-25 kg. These new particles move far away from each other, because they repel each other electrically (both are positively charged).

Because the calculations involve the small difference of (comparatively) large numbers, you need to keep seven significant figures in your calculations, and you need to use the more accurate value for the speed of light, 2.9979246e8 m/s. Choose all particles as the system.

Initial state: Original nucleus, at rest.

Final state: Alpha particle + new nucleus, far from each other. What is the sum of the kinetic energies of the alpha particle and the new nucleus?

Question

Tristian

Physics

18 December, 07:16

In a location in outer space far from all other objects, a nucleus whose mass is 3.969554 * 10-25 kg and that is initially at rest undergoes spontaneous alpha decay. The original nucleus disappears, and two new particles appear: a He-4 nucleus of mass 6.640678 * 10-27 kg (an alpha particle consisting of two protons and two neutrons) and a new nucleus of mass 3.902996 * 10-25 kg. These new particles move far away from each other, because they repel each other electrically (both are positively charged).

Because the calculations involve the small difference of (comparatively) large numbers, you need to keep seven significant figures in your calculations, and you need to use the more accurate value for the speed of light, 2.9979246e8 m/s. Choose all particles as the system.

Initial state: Original nucleus, at rest.

Final state: Alpha particle + new nucleus, far from each other. What is the sum of the kinetic energies of the alpha particle and the new nucleus?

+1

Answers (1)

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Madeleine Mcclain · Answer 1

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