An alpha-particle of energy 5 MeV is movingforward for a head on collision. The distance of closest approach from the nucleus of atomic numbe Z=50 is ... xx10^(-14)m (k=9xx10^(9)SI, c=1.6xx10^(-19)C),

An alpha-particle of energy 5 MeV is movingforward for a head on colli

1.	An alpha-particle of energy 5 MeV is movingforward for a head on collision. The distance of closest approach from the nucleus of atomic numbe Z=50 is ... xx10^(-14)m (k=9xx10^(9)SI, c=1.6xx10^(-19)C),
Answer» `0.72` `2.88` `1.44` `5.76` Solution :KINETIC energy of a-particle = Potential energy of a-particle at very large distance, `5xx10^(6)xx1.6xx10^(-19)=(k(2Ze^(2)))/(d)` `:.d=(k(2Ze^(2)))/(8XX10^(-13))` `:.d=(230.4xx10^(-28))/(8xx10^(-13))` `:.d=28.8xx10^(-5)m:d=2.88xx10^(-14)m`

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An alpha-particle of energy 5 MeV is movingforward for a head on collision. The distance of closest approach from the nucleus of atomic numbe Z=50 is ... xx10^(-14)m (k=9xx10^(9)SI, c=1.6xx10^(-19)C),

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