Question

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
A. `M_(2) = 2 M_(1)`
B. `M_(2) gt 2M_(1)`
C. `M_(2) lt 2M_(1)`
D. `M_(1) lt 10(m_(n) + m_(p))`

Answer 1

Correct Answer - C::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. `._(1.0)^(20)Ne` is made up of `10` protons plus `10` neutrons.
Therefore, mass of `._(10)^(20)Ne` nucleus,
`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)`
`implies 10 (m_(p)+m_(n)) gt M_(2)-M_(1) implies 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)`