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Let `m_(p)` be the mass of a poton , `M_(1)` the mass of a `_(10)^(20) Ne` nucleus and `M_(2)` the mass of a `_(20)^(40) Ca` nucleus . ThenA. `M_(2) = 2 M_(1)`B. `M_(2) gt 2M_(1)`C. `M_(2) lt 2M_(1)`D. `M_(1) lt 10(m_(n) + m_(p))` |
Answer» Correct Answer - C::D Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus is always less than the sum of masses of its constituent particles. `._(1.0)^(20)Ne` is made up of `10` protons plus `10` neutrons. Therefore, mass of `._(10)^(20)Ne` nucleus, `M_(1) lt 10 (m_(p)+m_(n))` Also, heavier the nucleus, more is the mass defect. Thus, `20 (m_(n)+m_(p))-M_(2) gt 10 (m_(p)+m_(n))-M_(1)` `implies 10 (m_(p)+m_(n)) gt M_(2)-M_(1) implies M_(2) lt M_(1)+10(m_(p)+m_(n))` Now since `M_(1) lt 10 (m_(p)+m_(n)) :. M_(2) lt 2M_(1)` |
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