The third line of the Balmer series spectrum of a hydrogen-like ion of atomic number `Z` equals to `108.5 nm`. The binding energy of the electron in the ground state of these ions is `E_(n)`. Then

Options

A. `Z = 2`
B. `E_(B) = 54.4 eV`
C. `Z = 3`
D. `E_(B) = 122.4 eV`

Explanation

To solve the problem, we need to determine the atomic number \( Z \) of the hydrogen-like ion and the binding energy of the electron in the ground state. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to transitions of electrons to the second energy level (n=2) from higher energy levels (n=3, 4, 5, ...). The third line of the Balmer series corresponds to a transition from \( n=5 \) to \( n=2 \).

### Step 2: Use the Balmer Formula
The wavelength \( \lambda \) of the emitted light can be calculated using the formula:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \]
where \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)), \( Z \) is the atomic number, and \( n \) is the principal quantum number of the higher energy level.

### Step 3: Substitute the Values
Given that the wavelength \( \lambda = 108.5 \, \text{nm} = 1.085 \times 10^{-7} \, \text{m} \) and \( n = 5 \):
\[ \frac{1}{\lambda} = \frac{1}{1.085 \times 10^{-7}} \approx 9.21 \times 10^6 \, \text{m}^{-1} \]
Substituting into the Balmer formula:
\[ 9.21 \times 10^6 = RZ^2 \left( \frac{1}{4} - \frac{1}{25} \right) \]
Calculating the right-hand side:
\[ \frac{1}{4} - \frac{1}{25} = \frac{25 - 4}{100} = \frac{21}{100} = 0.21 \]
Thus, the equation becomes:
\[ 9.21 \times 10^6 = RZ^2 \times 0.21 \]

### Step 4: Solve for \( Z^2 \)
Using \( R \approx 1.097 \times 10^7 \):
\[ 9.21 \times 10^6 = (1.097 \times 10^7) Z^2 \times 0.21 \]
\[ Z^2 = \frac{9.21 \times 10^6}{(1.097 \times 10^7) \times 0.21} \]
Calculating this gives:
\[ Z^2 \approx 4 \quad \Rightarrow \quad Z = 2 \]

### Step 5: Calculate the Binding Energy
The binding energy \( E_n \) of an electron in a hydrogen-like atom is given by:
\[ E_n = -\frac{13.6 Z^2}{n^2} \, \text{eV} \]
For the ground state (\( n = 1 \)):
\[ E_1 = -\frac{13.6 \times 2^2}{1^2} = -\frac{13.6 \times 4}{1} = -54.4 \, \text{eV} \]

### Final Answers
- The atomic number \( Z \) is \( 2 \).
- The binding energy of the electron in the ground state is \( -54.4 \, \text{eV} \).