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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)=2M_(1)`B. `M_(2) gt 2M_(1)`C. `M_(2) ltM_(1)`D. `M_(1) lt 10(m_(n)+m_(p))` |
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Answer» Correct Answer - 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. `._(10)^(20)Ne` is made up of 10 protons plus 10 neutrons. Therefore mass of `._(10)^(20) Ne` nucleaus `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)` or `10 (m_(p)+m_(n)) gt M_(2) -M_(1)` or `M_(2) lt M_(1)+10(m_(p)+m_(n))` Since `M_(1) gt 10(m_(p)+m_(n))` `:. M_(2) lt 2M_(1)` |
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