The wavelength of the first line of Lyman series in hydrogen atom is `1216`. The wavelength of the first line of Lyman series for `10` times ionized sodium atom will be added

Options

A. 1000 A
B. 100 A
C. `10 A`
D. `1A`

Answer

To solve the problem of finding the wavelength of the first line of the Lyman series for a 10 times ionized sodium atom (Na⁺¹⁰), we will follow these steps:

### Step 1: Understand the Lyman Series
The Lyman series corresponds to electronic transitions in hydrogen-like atoms where the electron falls from a higher energy level (n2) to the ground state (n1 = 1). The first line of the Lyman series corresponds to the transition from n2 = 2 to n1 = 1.

### Step 2: Identify the Given Data
- Wavelength of the first line of the Lyman series for hydrogen (λ_H) = 1216 Å (angstroms).
- Atomic number of hydrogen (Z_H) = 1.
- Atomic number of sodium (Z_Na) = 11.

### Step 3: Use the Rydberg Formula
The Rydberg formula for the wavelength of spectral lines in hydrogen-like atoms is given by:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- R is the Rydberg constant.
- Z is the atomic number.
- n1 and n2 are the principal quantum numbers of the lower and higher energy levels, respectively.

### Step 4: Set Up the Equation for Hydrogen and Sodium
For hydrogen:
\[ \frac{1}{\lambda_H} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

For sodium (Na⁺¹⁰):
\[ \frac{1}{\lambda_{Na}} = R \cdot 11^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

### Step 5: Relate the Wavelengths
Since the term \(\left( \frac{1}{1^2} - \frac{1}{2^2} \right)\) is the same for both hydrogen and sodium, we can relate the wavelengths as follows:

\[ \frac{\lambda_H}{\lambda_{Na}} = \frac{Z_{Na}^2}{Z_H^2} \]

### Step 6: Substitute the Values
Substituting the known values into the equation:

\[ \frac{1216 \, \text{Å}}{\lambda_{Na}} = \frac{11^2}{1^2} \]

\[ \frac{1216 \, \text{Å}}{\lambda_{Na}} = 121 \]

### Step 7: Solve for λ_Na
Rearranging the equation gives:

\[ \lambda_{Na} = \frac{1216 \, \text{Å}}{121} \]

Calculating this gives:

\[ \lambda_{Na} = 10 \, \text{Å} \]

### Final Answer
The wavelength of the first line of the Lyman series for the 10 times ionized sodium atom is **10 Å**.

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