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) gt 2M_(1)`
C. `M_(2) lt 2M_(1)`
D. `M_(1) lt10(m_(p) +m_(p))`

Answer

c,d

Explanation

Due to mass defect (which is finally responsible for the binding energy of the nucelus), mass of a nucleus is alwaya less than the sum of masses of its cosntituent particles. .`_(10)^(20)Ne` is made up of `10` protons plus `10` neutrons.
Therefore, mass of `._(10)^(20)Ne` nucleus,
Also, heavier the nucleus, more is the mass defect.
`20(m_(n) +m_(p))-M_(2) gt10 (m_(p) +m_(n))-M_(1)`
Thus, `10(m_(n)+m_(P))gtM_(2) -M_(1)`
or `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).`