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The ratio of the wavelengths for 2 rarr ...

The ratio of the wavelengths for `2 rarr 1` transition in `Li^(++), He^(+)` and `H` is

A

`1:2:3`

B

`1:4:9`

C

`2:9:36`

D

`3:2:1`

Text Solution

AI Generated Solution

The correct Answer is:
To find the ratio of the wavelengths for the \(2 \rightarrow 1\) transition in \(Li^{++}\), \(He^{+}\), and \(H\), we can use the Rydberg formula for the wavelengths of transitions in hydrogen-like atoms: \[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Where: - \(R\) is the Rydberg constant, - \(Z\) is the atomic number, - \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and upper energy levels, respectively. ### Step 1: Identify the transitions For the \(2 \rightarrow 1\) transition: - \(n_1 = 1\) - \(n_2 = 2\) ### Step 2: Write the formula for each ion 1. For \(H\) (Hydrogen, \(Z = 1\)): \[ \frac{1}{\lambda_H} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \] 2. For \(He^{+}\) (Helium, \(Z = 2\)): \[ \frac{1}{\lambda_{He}} = R \cdot 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \cdot 4 \left( 1 - \frac{1}{4} \right) = R \cdot 4 \cdot \frac{3}{4} = 3R \] 3. For \(Li^{++}\) (Lithium, \(Z = 3\)): \[ \frac{1}{\lambda_{Li}} = R \cdot 3^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \cdot 9 \left( 1 - \frac{1}{4} \right) = R \cdot 9 \cdot \frac{3}{4} = \frac{27R}{4} \] ### Step 3: Find the wavelengths Now, we can express the wavelengths: - \(\lambda_H = \frac{4}{3R}\) - \(\lambda_{He} = \frac{1}{3R}\) - \(\lambda_{Li} = \frac{4}{27R}\) ### Step 4: Find the ratios of the wavelengths Now, we can find the ratio of the wavelengths: \[ \lambda_H : \lambda_{He} : \lambda_{Li} = \frac{4}{3R} : \frac{4}{9R} : \frac{4}{27R} \] To simplify, we can multiply through by \(27R\) to eliminate the denominators: \[ \lambda_H : \lambda_{He} : \lambda_{Li} = 36 : 12 : 4 \] ### Step 5: Simplify the ratio Now, we can simplify this ratio: \[ 36 : 12 : 4 = 9 : 3 : 1 \] ### Final Ratio Thus, the final ratio of the wavelengths for the \(2 \rightarrow 1\) transition in \(Li^{++}\), \(He^{+}\), and \(H\) is: \[ 9 : 3 : 1 \]

To find the ratio of the wavelengths for the \(2 \rightarrow 1\) transition in \(Li^{++}\), \(He^{+}\), and \(H\), we can use the Rydberg formula for the wavelengths of transitions in hydrogen-like atoms: \[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Where: - \(R\) is the Rydberg constant, ...
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