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  • let be the mass of proton, <img alt="\mathrm{m_n }" src="https://entrancecorner.oncodecogs.com/gi

let \mathrm{m_p} be the mass of proton, \mathrm{m_n } the mass of neutron.\mathrm{ M_{\perp}} the mass of _{40}^{20}\textrm{Ne} nucleus and \mathrm{M}_2 the mass of _{20}^{10}\textrm{Ca} nucleus then -
 

Option: 1

\mathrm{M_2=2 \mathrm{M} }


Option: 2

\mathrm{M_2>2 M_1 }


Option: 3

\mathrm{M_2<2 M, }


Option: 4

None


Answers (1)

best_answer

\mathrm{_{10}Ne^{20}} is made up of 10 protons plus 10 neutrons.

Therefore mass of \mathrm{_{10}Ne^{20}} nucleus

\mathrm{M_1<10\left(m_p+m_n\right) }

Also heavier the nucleus, more is the mass defect.

\begin{aligned} & \mathrm{20\left(m_n+m_p\right)-m_2>10\left(m_p+m_n\right)-m_1} \\ & \mathrm{10\left(m_p+m_n\right)>m_2-m_1} \\ & \mathrm{m_0<m_1+10\left(m_p+m_n\right) }\end{aligned}

Now, since \mathrm{M,<10\left(m_p+m_n\right) }

\mathrm{\therefore M_2<\& M_1 }
 

Posted by

Pankaj

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