**Problem Description:**Two beta-minus particles, each with 75 keV of kinetic energy, undergo Bremsstrahlung interactions. One beta-minus particle has 35 keV after the interaction, and the other has 61 keV after the interaction. What is the Bremsstrahlung X-ray energy in each case? Which of the two beta-minus particles came closest to the nucleus?**Solution Explanation:**Bremsstrahlung interaction involves the emission of X-rays due to the deceleration of beta particles when they come close to atomic nuclei. The difference in kinetic energy before and after the interaction determines the X-ray energy emitted.1. **First Beta-minus Particle:** - Initial Kinetic Energy: 75 keV - Final Kinetic Energy: 35 keV - Bremsstrahlung X-ray Energy = Initial Kinetic Energy - Final Kinetic Energy = 75 keV - 35 keV = 40 keV2. **Second Beta-minus Particle:** - Initial Kinetic Energy: 75 keV - Final Kinetic Energy: 61 keV - Bremsstrahlung X-ray Energy = Initial Kinetic Energy - Final Kinetic Energy = 75 keV - 61 keV = 14 keV**Conclusion:**The first beta-minus particle, which had a higher Bremsstrahlung X-ray energy of 40 keV, came closest to the nucleus as it lost more kinetic energy compared to the second particle.

Two beta-minus particles, each with 75 keV of kinetic energy, undergo Bremsstrahlung interactions. One beta-minus particle has 35 kev after the interaction, and the other has 61 keV after the interaction. What is the Bremsstrahlung x-ray energy in each case? Which of the two beta-minus particles came closest to the nucleus?

Two beta-minus particles, each with 75 keV of kinetic energy, undergo Bremsstrahlung interactions. One beta-minus particle has 35 kev after the interaction, and the other has 61 keV after the interaction. What is the Bremsstrahlung x-ray energy in each case? Which of the two beta-minus particles came closest to the nucleus?

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