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) ltM_(1)`
D. `M_(1) lt 10(m_(n)+m_(p))`

Answer

D

Explanation

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)`