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Nuclei with magic number of proton $\mathrm{Z}=2,8,20,28,50$, 52 and magic number of neutrons $\mathrm{N}=2,8,20,28,50,82$ and 126 are found to be very stable
(i) Verify this by caculating the proton. Separation energy $S_p$ for ${ }^{120} \mathrm{Sn}(\mathrm{Z}=50)$ and ${ }^{121} \mathrm{Sb}(\mathrm{Z}=51)$.
The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by
$$
\begin{aligned}
&\mathrm{S}_{\mathrm{p}}=\left(\mathrm{M}_{\mathrm{Z}-1{ }^1 \mathrm{~N}}+\mathrm{M}_{\mathrm{H}}-\mathrm{M}_{\mathrm{Z}, \mathrm{N}}\right) \mathrm{c}^2 . \\
&\text { Given, }{ }^{19} \ln =118.9058 \mathrm{u}, \\
&{ }^{120} \mathrm{Sn}=199.902199 \mathrm{u}, \\
&{ }^{121} \mathrm{Sb}=120.903824 \mathrm{u},{ }^1 \mathrm{H}=1.0078252 \mathrm{u} .
\end{aligned}
$$
(ii) What does the existence of magic number indicate?
(i) Verify this by caculating the proton. Separation energy $S_p$ for ${ }^{120} \mathrm{Sn}(\mathrm{Z}=50)$ and ${ }^{121} \mathrm{Sb}(\mathrm{Z}=51)$.
The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by
$$
\begin{aligned}
&\mathrm{S}_{\mathrm{p}}=\left(\mathrm{M}_{\mathrm{Z}-1{ }^1 \mathrm{~N}}+\mathrm{M}_{\mathrm{H}}-\mathrm{M}_{\mathrm{Z}, \mathrm{N}}\right) \mathrm{c}^2 . \\
&\text { Given, }{ }^{19} \ln =118.9058 \mathrm{u}, \\
&{ }^{120} \mathrm{Sn}=199.902199 \mathrm{u}, \\
&{ }^{121} \mathrm{Sb}=120.903824 \mathrm{u},{ }^1 \mathrm{H}=1.0078252 \mathrm{u} .
\end{aligned}
$$
(ii) What does the existence of magic number indicate?
Solution:
2132 Upvotes
Verified Answer
(i) The proton separation energy is
So, $\mathrm{S}_{\mathrm{p}}$ for 50
$\mathrm{Sn}^{120}=\left(\mathrm{M}_{119.70}+\mathrm{M}_{\mathrm{H}}-\mathrm{M}_{120,70}\right) \mathrm{c}^2$ $=(118.9058+1.0078252-119.902199) \mathrm{c}^2$
$\mathrm{S}_{\mathrm{p}}$ for $\mathrm{Sn}^{120}=0.0114362 \mathrm{c}^2$
at $z=51, z-1=50$ for $\mathrm{Sn}$
Similarly for $\mathrm{S}_{\mathrm{p}}$ of
$$
\begin{aligned}
&\mathrm{S}_{\mathrm{n}} / \mathrm{SpSb}=\left(\mathrm{M}_{120,70}+\mathrm{M}_{\mathrm{H}}-\mathrm{M}_{121,70}\right) \mathrm{c}^2 \\
&=\left(199.902199+1.007825^2-120.903822\right) \mathrm{c}^2 \\
&=0.0059912 \mathrm{c}^2
\end{aligned}
$$
Since, $\mathrm{S}_{\mathrm{psn}}>\mathrm{S}_{\mathrm{pSb}}$, Sn nucleus is more stable than $\mathrm{Sb}$ nucleus.
(ii) The existence of magic numbers indicates that the shell structure of nucleus similar to the shell structure of an atom. This also explains the peaks in binding energy/nucleon curve.
So, $\mathrm{S}_{\mathrm{p}}$ for 50
$\mathrm{Sn}^{120}=\left(\mathrm{M}_{119.70}+\mathrm{M}_{\mathrm{H}}-\mathrm{M}_{120,70}\right) \mathrm{c}^2$ $=(118.9058+1.0078252-119.902199) \mathrm{c}^2$
$\mathrm{S}_{\mathrm{p}}$ for $\mathrm{Sn}^{120}=0.0114362 \mathrm{c}^2$
at $z=51, z-1=50$ for $\mathrm{Sn}$
Similarly for $\mathrm{S}_{\mathrm{p}}$ of
$$
\begin{aligned}
&\mathrm{S}_{\mathrm{n}} / \mathrm{SpSb}=\left(\mathrm{M}_{120,70}+\mathrm{M}_{\mathrm{H}}-\mathrm{M}_{121,70}\right) \mathrm{c}^2 \\
&=\left(199.902199+1.007825^2-120.903822\right) \mathrm{c}^2 \\
&=0.0059912 \mathrm{c}^2
\end{aligned}
$$
Since, $\mathrm{S}_{\mathrm{psn}}>\mathrm{S}_{\mathrm{pSb}}$, Sn nucleus is more stable than $\mathrm{Sb}$ nucleus.
(ii) The existence of magic numbers indicates that the shell structure of nucleus similar to the shell structure of an atom. This also explains the peaks in binding energy/nucleon curve.
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