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If the energy of the α particle emitted by 231Am is 5.48 MeV, find the distance of closest approach between the α particle and 197Au nucleus (in 10–4 m)
    Correct answer is '2'. Can you explain this answer?
    Verified Answer
    If the energy of the α particle emitted by 231Am is 5.48 MeV, fi...
    K.E, = 5.48 MeV
    = 5.48 × 106 × 1.6 × 10-19 J


    The correct answer is: 2
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    If the energy of the α particle emitted by 231Am is 5.48 MeV, fi...
    If the energy of the system is conserved, it means that the total amount of energy in the system remains constant over time. This means that energy cannot be created or destroyed, but it can be transferred or converted from one form to another. In other words, the total energy of the system before and after any processes or interactions within the system will be the same.
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    If the energy of the α particle emitted by 231Am is 5.48 MeV, find the distance of closest approach between the α particle and 197Au nucleus (in 10–4 m)Correct answer is '2'. Can you explain this answer?
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