Nucleus of 92U238 at rest emits an alpha particle with velocity of 1.4...
Explanation:
Conservation of Momentum:
- When the nucleus of 92U238 emits an alpha particle, according to the law of conservation of momentum, the total momentum before the emission must be equal to the total momentum after the emission.
- The momentum of the alpha particle is given by p = mv, where m is the mass of the alpha particle and v is its velocity.
- The momentum of the remaining nucleus can be calculated as the initial momentum of the system minus the momentum of the alpha particle.
Calculation:
- Let's assume the initial momentum of the system is 0 since the nucleus is at rest.
- After emitting the alpha particle, the momentum of the alpha particle is given by p_alpha = m_alpha * v_alpha, where m_alpha is the mass of the alpha particle and v_alpha is its velocity.
- The momentum of the remaining nucleus is then given by p_nucleus = -p_alpha (since momentum is conserved).
Velocity of the Remaining Nucleus:
- The velocity of the remaining nucleus can be calculated by dividing the momentum of the nucleus by its mass.
- Since momentum is a vector quantity, the velocity of the remaining nucleus will be in the opposite direction to that of the alpha particle.
Conclusion:
- The velocity of the remaining nucleus after emitting an alpha particle with a velocity of 1.4 × 10^7 m/s can be calculated using the conservation of momentum principle, and it will be in the opposite direction with a magnitude determined by the mass of the nucleus and the momentum of the alpha particle.
Nucleus of 92U238 at rest emits an alpha particle with velocity of 1.4...
Momentum conservation se ho jayga. ...
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