The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of `108.5mm`. The ground state energy of an electron of this ion will e

Options

A. `3.4 eV`
B. `13.6 eV`
C. `54.4 eV`
D. `122.4 eV`

Explanation

To solve the problem, we need to find the ground state energy of an electron in an ion that is equivalent to the hydrogen atom, given that the third line of the Balmer series has a wavelength of 108.5 nm.

### Step-by-Step Solution:

**Step 1: Understand the Balmer Series**
The Balmer series corresponds to transitions where an electron falls to the second energy level (n=2) from higher energy levels (n=3, n=4, n=5, ...). The third line of the Balmer series corresponds to the transition from n=5 to n=2.

**Step 2: Use the Rydberg Formula**
The Rydberg formula for the wavelength (λ) of the emitted light during such transitions 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 (\( R \approx 1.1 \times 10^7 \, \text{m}^{-1} \))
- \( Z \) is the atomic number of the ion
- \( n_1 \) is the lower energy level (2 for Balmer series)
- \( n_2 \) is the higher energy level (5 for the third line)

**Step 3: Convert Wavelength to Meters**
Given the wavelength \( \lambda = 108.5 \, \text{nm} \), we convert it to meters:

\[ \lambda = 108.5 \times 10^{-9} \, \text{m} \]

**Step 4: Substitute Values into the Rydberg Formula**
Substituting \( n_1 = 2 \), \( n_2 = 5 \), and \( \lambda = 108.5 \times 10^{-9} \, \text{m} \) into the Rydberg formula:

\[ \frac{1}{108.5 \times 10^{-9}} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \]

Calculating the right side:

\[ \frac{1}{2^2} = \frac{1}{4}, \quad \frac{1}{5^2} = \frac{1}{25} \]

\[ \frac{1}{4} - \frac{1}{25} = \frac{25 - 4}{100} = \frac{21}{100} = 0.21 \]

Thus, we have:

\[ \frac{1}{108.5 \times 10^{-9}} = RZ^2 (0.21) \]

**Step 5: Solve for Z**
Now, rearranging the equation to solve for \( Z^2 \):

\[ Z^2 = \frac{1}{108.5 \times 10^{-9} \times R \times 0.21} \]

Substituting \( R = 1.1 \times 10^7 \):

\[ Z^2 = \frac{1}{108.5 \times 10^{-9} \times 1.1 \times 10^7 \times 0.21} \]

Calculating this gives us \( Z^2 \). After performing the calculations, we find \( Z = 2 \).

**Step 6: Calculate the Ground State Energy**
The ground state energy \( E \) of an electron in a hydrogen-like ion is given by:

\[ E = -\frac{13.6 Z^2}{n^2} \]

For the ground state, \( n = 1 \):

\[ E = -\frac{13.6 \times 2^2}{1^2} = -\frac{13.6 \times 4}{1} = -54.4 \, \text{eV} \]

### Final Answer:
The ground state energy of an electron in this ion is \( -54.4 \, \text{eV} \).

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