Modern Physics Ques 196
- Let $m _p$ be the mass of proton, $m _n$ the mass of neutron. $M _1$ the mass of ${ } _{10}^{20} Ne$ nucleus and $M _2$ the mass of ${ } _{20}^{40} Ca$ nucleus. Then
$(1998,2 M)$
(a) $M _2=2 M _1$
(b) $M _2>2 M _1$
(c) $M _2<2 M _1$
(d) $M _1<10\left(m _n+m _p\right)$
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Answer:
Correct Answer: 196.(c,d)
Solution:
Formula:
- 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 nucleus, $M _1<10\left(m _p+m _n\right)$.
Also, heavier the nucleus, more is the mass defect.
Thus, $20\left(m _n+m _p\right)-M _2>10\left(m _p+m _n\right)-M _1$
$ \begin{array}{ll} \text { or } & 10\left(m _p+m _n\right)>M _2-M _1 \\ \text { or } & M _2<M _1+10\left(m _p+m _n\right) \\ \text { Now since } & M _1<10\left(m _p+m _n\right) \\ \therefore & M _2<2 M _1 \end{array} $
BETA
