Considering the Bohr model of hydrogen like atoms, the ratio of the radius $5^{\text {th }}$ orbit of the electron in $\mathrm{Li}^{2+}$ and $\mathrm{He}^{+}$is
Considering the Bohr model of hydrogen like atoms, the ratio of the radius $5^{\text {th }}$ orbit of the electron in $\mathrm{Li}^{2+}$ and $\mathrm{He}^{+}$is
Show Hint
- $\frac{3}{2}$
- $\frac{4}{9}$
- $\frac{9}{4}$
- $\frac{2}{3}$
The Correct Option is D
Approach Solution - 1
2. Radius of the $5^{\text{th}}$ orbit for $\mathrm{He}^{+}$: \[ r_{5} = \frac{5^2}{2} a_0 \]
3. Ratio of the radii: \[ \frac{r_{\mathrm{Li}^{2+}}}{r_{\mathrm{He}^{+}}} = \frac{2}{3} \] Therefore, the correct answer is (4) $\frac{2}{3}$.
Approach Solution -2
We are asked to find the ratio of the radius of the 5th orbit of an electron in \( \mathrm{Li}^{2+} \) and \( \mathrm{He}^{+} \) ions according to the Bohr model.
Concept Used:
In the Bohr model, the radius of the \( n^{\text{th}} \) orbit for a hydrogen-like atom is given by:
\[ r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2 Z} = a_0 \frac{n^2}{Z} \]
where:
- \( a_0 = 0.529 \times 10^{-10} \, \text{m} \) (Bohr radius)
- \( n \) = principal quantum number
- \( Z \) = atomic number of the element
Step-by-Step Solution:
Step 1: Write the expression for the radius of the 5th orbit for both ions.
\[ r_{5,\mathrm{Li}^{2+}} = a_0 \frac{5^2}{Z_{\mathrm{Li}}} \] \[ r_{5,\mathrm{He}^{+}} = a_0 \frac{5^2}{Z_{\mathrm{He}}} \]
Step 2: Substitute the atomic numbers:
- For \( \mathrm{Li}^{2+} \), \( Z = 3 \)
- For \( \mathrm{He}^{+} \), \( Z = 2 \)
Step 3: Substitute into the formula and simplify.
\[ r_{5,\mathrm{Li}^{2+}} = a_0 \frac{25}{3} \] \[ r_{5,\mathrm{He}^{+}} = a_0 \frac{25}{2} \]
Step 4: Take the ratio.
\[ \frac{r_{5,\mathrm{Li}^{2+}}}{r_{5,\mathrm{He}^{+}}} = \frac{\frac{25}{3}}{\frac{25}{2}} = \frac{2}{3} \]
Final Computation & Result:
The ratio of the radii of the 5th orbit is:
\[ \boxed{\frac{r_{5,\mathrm{Li}^{2+}}}{r_{5,\mathrm{He}^{+}}} = \frac{2}{3}} \]
Final Answer: \( \dfrac{2}{3} \)
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