Question:
The radius in the hydrogen atom in the ground state is $0.53$ \AA. The radius of $Li^{+2}$ ion (atomic number $= 3$) in a similar state is \ldots..
The radius in the hydrogen atom in the ground state is $0.53$ \AA. The radius of $Li^{+2}$ ion (atomic number $= 3$) in a similar state is \ldots..
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To quickly solve Bohr radius problems, remember that as the nuclear charge ($Z$) increases, the nucleus pulls the electron closer, making the radius smaller. Since Lithium has 3 protons ($Z=3$) compared to Hydrogen's 1, its ground state radius must be exactly one-third of Hydrogen's.
Updated On: May 12, 2026
- $0.53 \text{\AA}$
- $1.06 \text{\AA}$
- $0.17 \text{\AA}$
- $0.265 \text{\AA}$
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The Correct Option is C
Solution and Explanation
Concept:
According to Bohr's model of the atom, the radius of the $n^{th}$ orbit for a hydrogen-like species (single-electron systems like $H$, $He^+$, $Li^{2+}$, etc.) is given by the formula:
$$r_n = a_0 \frac{n^2}{Z}$$
Where:
• $r_n$ is the radius of the $n^{th}$ orbit.
• $a_0$ is the Bohr radius (radius of H-atom in ground state, $\approx 0.529 \text{\AA}$ or $0.53 \text{\AA}$).
• $n$ is the principal quantum number (orbit number).
• $Z$ is the atomic number of the element.
Step 1: Identify the given values for Hydrogen and Lithium. For the Hydrogen atom in the ground state:
• Principal quantum number ($n$) = $1$ (since it is the ground state).
• Atomic number ($Z_H$) = $1$.
• Radius ($r_H$) = $0.53 \text{\AA}$. For the $Li^{+2}$ ion in a similar state:
• Principal quantum number ($n$) = $1$ (as it is in a "similar state" i.e., ground state).
• Atomic number for Lithium ($Z_{Li}$) = $3$.
Step 2: Establishing the relationship between radii. Using the formula $r \propto \frac{n^2}{Z}$, and since $n$ is constant ($n=1$) for both cases, we can state that:
$$r \propto \frac{1}{Z}$$
Therefore, the ratio of the radius of the $Li^{+2}$ ion to the radius of the Hydrogen atom is: $$\frac{r_{Li^{+2}}}{r_H} = \frac{Z_H}{Z_{Li}}$$
Step 3: Calculating the final radius.
Substitute the known values into the ratio: $$\frac{r_{Li^{+2}}}{0.53} = \frac{1}{3}$$
Now, solve for $r_{Li^{+2}}$:
$$r_{Li^{+2}} = \frac{0.53}{3}$$
$$r_{Li^{+2}} = 0.17666... \text{\AA}$$
Rounding to two decimal places as per the options provided:
$$r_{Li^{+2}} \approx 0.17 \text{\AA}$$
$$r_n = a_0 \frac{n^2}{Z}$$
Where:
• $r_n$ is the radius of the $n^{th}$ orbit.
• $a_0$ is the Bohr radius (radius of H-atom in ground state, $\approx 0.529 \text{\AA}$ or $0.53 \text{\AA}$).
• $n$ is the principal quantum number (orbit number).
• $Z$ is the atomic number of the element.
Step 1: Identify the given values for Hydrogen and Lithium. For the Hydrogen atom in the ground state:
• Principal quantum number ($n$) = $1$ (since it is the ground state).
• Atomic number ($Z_H$) = $1$.
• Radius ($r_H$) = $0.53 \text{\AA}$. For the $Li^{+2}$ ion in a similar state:
• Principal quantum number ($n$) = $1$ (as it is in a "similar state" i.e., ground state).
• Atomic number for Lithium ($Z_{Li}$) = $3$.
Step 2: Establishing the relationship between radii. Using the formula $r \propto \frac{n^2}{Z}$, and since $n$ is constant ($n=1$) for both cases, we can state that:
$$r \propto \frac{1}{Z}$$
Therefore, the ratio of the radius of the $Li^{+2}$ ion to the radius of the Hydrogen atom is: $$\frac{r_{Li^{+2}}}{r_H} = \frac{Z_H}{Z_{Li}}$$
Step 3: Calculating the final radius.
Substitute the known values into the ratio: $$\frac{r_{Li^{+2}}}{0.53} = \frac{1}{3}$$
Now, solve for $r_{Li^{+2}}$:
$$r_{Li^{+2}} = \frac{0.53}{3}$$
$$r_{Li^{+2}} = 0.17666... \text{\AA}$$
Rounding to two decimal places as per the options provided:
$$r_{Li^{+2}} \approx 0.17 \text{\AA}$$
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