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For which of the following, the radius will be the same for the hydrogen atom having n=1?
(A$H{{e}^{+}}$, $\text{n}=\,2$
(B)\[\text{L}{{\text{i}}^{2+}}\], \[n\,=\,2\]
(C)\[\text{B}{{\text{e}}^{3+}}\] , \[n\,=\,2\]
(D) None


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Last updated date: 13th Jun 2024
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Answer
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Hint: Calculation of the radius of an atom is the application of Bohr’s model, which is applicable for hydrogen and hydrogen like single electron containing species such as \[\text{L}{{\text{i}}^{2+}}\], \[\text{B}{{\text{e}}^{3+}}\] ion.

Complete Step by step solution:
Radius or orbital (shell) of a single electron containing species is calculated by
\[r\,=\,\dfrac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}m{{e}^{2}}}\times \dfrac{1}{Z}\]
 In it\[h\],\[\pi \] \[m\]and \[e\] are constant, so after substituting the value we get
\[\text{r}\,\,=\,0.529\times \,\dfrac{{{\text{n}}^{\text{2}}}}{\text{Z}}{{\text{A}}^{\text{o}}}\,......(1)\]
Where $\text{n}$ = no of orbit, and Z = atomic number
For hydrogen atom, the value of radius after putting n = 1 and Z = 1 in the equation (1)
 \[\begin{align}
& \text{r}\,\,=\,0.529\times \,\dfrac{{{\text{n}}^{\text{2}}}}{\text{Z}}{{\text{A}}^{\text{o}}}......\left( 1 \right) \\
 & \text{r}\,\,=\,0.529\times \,\dfrac{{{1}^{\text{2}}}}{1}{{\text{A}}^{\text{o}}} \\
 & \text{r}\,\,=\,\,0.529\,{{\text{A}}^{\text{o}}} \\
\end{align}\]
(A) $H{{e}^{+}}$Ion has only one electron but it has two protons in the nucleus, hence its electron feels three times more attraction from the nucleus in comparison to the hydrogen atom. Thus the radius of the ion for $\text{n}=\,2$ will be
After putting these value $\text{n}=\,2$ and \[\text{Z}\,\text{= 2}\] in equation (1)
\[\begin{align}
& \text{r}\,\,=\,0.529\times \,\dfrac{{{\text{n}}^{\text{2}}}}{\text{Z}}{{\text{A}}^{\text{o}}}......\left( 1 \right) \\
 & \text{r}\,\,=\,0.529\times \,\dfrac{{{2}^{\text{2}}}}{2}{{\text{A}}^{\text{o}}} \\
 & \text{r}\,\,=\,\,2\,\times 0.529\,{{\text{A}}^{\text{o}}} \\
\end{align}\]
\[\text{r}\,=\,1.05\,{{A}^{\circ }}\]
(B) \[L{{i}^{2+}}\]Ion has only one electron but it has three protons in the nucleus. So radius of\[\text{L}{{\text{i}}^{2+}}\]ion \[n\,=\,2\]and\[\text{Z}\,\text{= 3}\]after putting these values in the equation (1) we get
\[\begin{align}
& \text{r}\,\,=\,0.529\times \,\dfrac{{{\text{n}}^{\text{2}}}}{\text{Z}}{{\text{A}}^{\text{o}}}......\left( 1 \right) \\
& \text{r}\,\,=\,0.529\times \,\dfrac{{{2}^{\text{2}}}}{3}{{\text{A}}^{\text{o}}} \\
& r\,\,=\,\,0.529\times \,\dfrac{4}{3}{{\text{A}}^{\text{o}}}\,\, \\
\end{align}\]
\[\text{r}\,\,=\,\,0.235\,{{\text{A}}^{\text{o}}}\,\,\]
(C) \[\text{B}{{\text{e}}^{3+}}\]Ion has only one electron but it has four protons in the nucleus. So radius of \[\text{B}{{\text{e}}^{3+}}\]for its second orbital \[n\,=\,2\]and\[\text{Z}\,\text{= 4}\] after putting these values on equation (1) we get
 \[\begin{align}
& \text{r}\,\,=\,0.529\times \,\dfrac{{{\text{n}}^{\text{2}}}}{\text{Z}}{{\text{A}}^{\text{o}}}......\left( 1 \right) \\
& \text{r}\,\,=\,0.529\times \,\dfrac{{{2}^{\text{2}}}}{4}{{\text{A}}^{\text{o}}} \\
& \text{r}\,\,=\,\,0.529\,{{\text{A}}^{\text{o}}} \\
\end{align}\]

So the option (C) will be the correct option.

Note: radius of an atomic shell is directly proportional to the nth number of shell and inversely proportional to the atomic number or number of protons in the atom. So for a single electron species the size of the first shell will be the smallest and size of last shall be highest.