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 . Then

Options

A. `M_(2)=2M_(1)`
B. `M_(2)gt2M_(1)`
C. `M_(2)lt2M_(1)`
D. `M_(1) lt10(m_(n)+m_(p))`

Answer

C, D

Explanation

On account of mass defect, `M_(1)lt10(m_(n)+m_(p))` and `M_(2)lt20(m_(p)+m_(n))`
As B.E. in case of `._(20)Ca^(40)` is more than B.E. in case of `._(10)Ca^(20)`, therefore, `M_(2)lt2M_(1)`