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pochemuha
2 years ago
6

A nucleus whose mass is 3.499612×10^(−25) kg undergoes spontaneous alpha decay. The original nucleus disappears and there appear

two new particles:
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 3.433132×10^(−25) kg (note that the new nucleus has less mass than the original nucleus, and it has two fewer protons and two fewer neutrons).
When the alpha particle has moved far away from the new nucleus (so the electric interactions are negligible), what is the combined kinetic energy of the alpha particle and new nucleus?
Physics
1 answer:
Elanso [62]2 years ago
5 0

Answer:

The sum of the kinetic energies of the alpha particle and the new nucleus = (6.5898 × 10⁻¹³) J

Explanation:

Old nucleus ---> New nucleus + alpha particle.

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 = (3.499612 × 10⁻²⁵) kg

Mass of new nucleus = mₙ = (3.433132 × 10⁻²⁵) kg

Mass of alpha particle = mₐ = (6.640678 × 10⁻²⁷) kg

Speed of light = c = (3.0 × 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) = (3.0 × 10⁸)² [(3.499612 × 10⁻²⁵) - (3.433132 × 10⁻²⁵) - (6.640678 × 10⁻²⁷)]

(kₙ + kₐ) = (9.0 × 10¹⁶)(0.00007322 × 10⁻²⁵) = (6.5898 × 10⁻¹³) J

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Answer: 98.

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By substituting l = 0.28m into the equation above, we have that

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Sound from the string in the pipe is produced at the second overtone ( which is the third harmonic).

Therefore the length of air in the pipe at second overtone ( third harmonic) is given below as

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V = fpλp

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As stated in the question, the frequency of sound in pipe is the same as that in string (fp = fs = f) , thus to get the velocity of sound wave in string we use the same formulae of

v = fλ

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