The ground state energy of hydrogen atom is -13.6eV. If an electron makes a transition form an energy level -0.85 eV to -1.51 eV, calculate the wavelength of spectral line emitted. To which series of hydrogen spectrum does this wavelength belongs?

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the energy levels
We know that the energy of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

We need to find the principal quantum numbers (n) corresponding to the given energy levels:
- For \( E = -0.85 \, \text{eV} \)
- For \( E = -1.51 \, \text{eV} \)

### Step 2: Calculate n for -0.85 eV
Using the formula:
\[ -0.85 = -\frac{13.6}{n^2} \]
Rearranging gives:
\[ n^2 = \frac{13.6}{0.85} \]
Calculating \( n^2 \):
\[ n^2 \approx 16 \]
Thus, \( n = 4 \).

### Step 3: Calculate n for -1.51 eV
Using the same formula:
\[ -1.51 = -\frac{13.6}{n^2} \]
Rearranging gives:
\[ n^2 = \frac{13.6}{1.51} \]
Calculating \( n^2 \):
\[ n^2 \approx 9 \]
Thus, \( n = 3 \).

### Step 4: Determine the transition
The electron transitions from \( n = 4 \) to \( n = 3 \).

### Step 5: Calculate the change in energy (ΔE)
The change in energy during the transition is given by:
\[ \Delta E = E_4 - E_3 \]
Substituting the values:
\[ \Delta E = (-0.85) - (-1.51) = 0.66 \, \text{eV} \]

### Step 6: Convert ΔE to Joules
To calculate the wavelength, we need to convert the energy from eV to Joules:
\[ \Delta E \text{ (in Joules)} = 0.66 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.056 \times 10^{-19} \, \text{J} \]

### Step 7: Calculate the wavelength (λ)
Using the formula:
\[ \lambda = \frac{hc}{\Delta E} \]
Where:
- \( h = 6.63 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \)

Substituting the values:
\[ \lambda = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{1.056 \times 10^{-19}} \]
Calculating \( \lambda \):
\[ \lambda \approx 1.875 \times 10^{-7} \, \text{m} = 18750 \, \text{Å} \]

### Step 8: Identify the series
The transition from \( n = 4 \) to \( n = 3 \) corresponds to the Balmer series of the hydrogen spectrum.

### Final Answer
The wavelength of the spectral line emitted is approximately \( 18750 \, \text{Å} \) and it belongs to the Balmer series.

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