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a polonium isotope with an atomic mass of 207.981246 u207.981246 u undergoes alpha decay, resulting in a daughter isotope with an atomic mass of 203.973044 u.203.973044 u. ignoring any recoil of the daughter, find the kinetic energy of the emitted alpha particle in megaelectronvolts (mev).

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bharathparasad577

The kinetic energy of the emitted alpha particle is 9.53MeV.

The kinetic energy that the alpha particle has emitted, is the energy in excess after removing the resting energy of the atoms and the helium nucleus that forms the alpha particle

Since energy and masses are related and cannot be

         m₀ c² =  c² + m_He c²+ K

         K = c² (m₀ - m_{f} - m_He)

the mass of the Helium atom is 4 u

          K = (3 10⁸)² (211,988868 -207.976652 - 4,002) 1,661 10⁻²⁷

          K = 14,949 10⁻¹¹ (0.0102)

         K = 1,527 10⁻¹² J

let's reduce 1 J = 6,242 10¹² MeV

          K = 9.53 MeV

Therefore, The kinetic energy of the emitted alpha particle is 9.53MeV.

To know more about energy, refer: https://brainly.com/question/15563159

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