Find the binding anergy of an electron in the ground state of a hydrogen like atom in whose spectrum the thrid Balmer line is equal to 108.5 mm.

Options

A. `54.4 eV`
B. `13.6 eV`
C. `112.4 eV`
D. None of these

Explanation

To find the binding energy of an electron in the ground state of a hydrogen-like atom, given that the third Balmer line has a wavelength of 108.5 nm, we can follow these steps:

### Step 1: Understand the Balmer series
The Balmer series corresponds to transitions of electrons in a hydrogen-like atom from higher energy levels (n ≥ 2) to the n = 2 level. The third line in the Balmer series corresponds to a transition from n = 5 to n = 2.

### Step 2: Calculate the energy of the photon emitted
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \) eV·s),
- \( c \) is the speed of light (\( 3 \times 10^8 \) m/s),
- \( \lambda \) is the wavelength in meters.

Convert the wavelength from nm to meters:
\[ \lambda = 108.5 \, \text{nm} = 108.5 \times 10^{-9} \, \text{m} \]

Now, substituting the values:
\[ E = \frac{(4.135667696 \times 10^{-15} \, \text{eV·s})(3 \times 10^8 \, \text{m/s})}{108.5 \times 10^{-9} \, \text{m}} \]

Calculating this gives:
\[ E \approx 11.4 \, \text{eV} \]

### Step 3: Relate the energy to the binding energy
The energy of the photon corresponds to the difference in energy levels:
\[ E = E_2 - E_5 \]
Using the formula for the energy levels of a hydrogen-like atom:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
where \( Z \) is the atomic number.

For the transition from n = 5 to n = 2:
\[ E = \left(-\frac{Z^2 \cdot 13.6}{2^2}\right) - \left(-\frac{Z^2 \cdot 13.6}{5^2}\right) \]
\[ E = Z^2 \cdot 13.6 \left(\frac{1}{4} - \frac{1}{25}\right) \]

### Step 4: Simplify the expression
Calculating the fractions:
\[ \frac{1}{4} = \frac{25}{100} \]
\[ \frac{1}{25} = \frac{4}{100} \]
Thus,
\[ \frac{1}{4} - \frac{1}{25} = \frac{25 - 4}{100} = \frac{21}{100} \]

Now substituting this back into the energy equation:
\[ E = Z^2 \cdot 13.6 \cdot \frac{21}{100} \]

### Step 5: Solve for the binding energy
We know \( E \approx 11.4 \, \text{eV} \):
\[ 11.4 = Z^2 \cdot 13.6 \cdot \frac{21}{100} \]
Rearranging gives:
\[ Z^2 = \frac{11.4 \cdot 100}{13.6 \cdot 21} \]

Calculating \( Z^2 \):
\[ Z^2 \approx \frac{1140}{285.6} \approx 4 \]
Thus, \( Z \approx 2 \).

### Step 6: Calculate the binding energy
Now we can find the binding energy:
\[ \text{Binding Energy} = 13.6 Z^2 \]
Substituting \( Z = 2 \):
\[ \text{Binding Energy} = 13.6 \cdot 2^2 = 13.6 \cdot 4 = 54.4 \, \text{eV} \]

### Final Answer
The binding energy of the electron in the ground state of the hydrogen-like atom is approximately **54.4 eV**.