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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) lt 2M_(1)`D. `M_(1) lt 10(m_(n) + m_(p))` |
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Answer» Correct Answer - C::D KEY CONCEPT : Due to mass defact (which is finally reponsible of the binding energy of the mucleus ) mass of a mecleus is always less then the sum of masses of its constiluent particles `_(10)^(20)` Ne is made up of `10` protons `10` neutron therefore , mass of`_(10)^(20)` Ne nucleus `m_(1) lt 10(m_(p) + m_(n))` |
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