The energy of stable states of the hydrogen atom is given by `E_(n)=-2.18xx10^(-8)//n^(2)[J]` where n denotes the principal quantum number.
In what spectral range is the Lyman series lying?

Answer

To determine the spectral range of the Lyman series for the hydrogen atom, we will follow these steps:

### Step 1: Understand the Lyman Series
The Lyman series corresponds to electronic transitions in a hydrogen atom where the electron falls to the n=1 energy level (ground state). The transitions can occur from higher energy levels (n=2, n=3, n=4, etc.) down to n=1.

### Step 2: Calculate the Energy Levels
The energy of the stable states of the hydrogen atom is given by the formula:
\[ E_n = -\frac{2.18 \times 10^{-8}}{n^2} \text{ J} \]
We need to calculate the energy for n=1 (ground state) and for n approaching infinity (ionization).

- For n=1:
\[ E_1 = -\frac{2.18 \times 10^{-8}}{1^2} = -2.18 \times 10^{-8} \text{ J} \]

- For n approaching infinity (ionization):
\[ E_{\infty} = -\frac{2.18 \times 10^{-8}}{\infty^2} = 0 \text{ J} \]

### Step 3: Calculate the Energy Difference
The energy difference (ΔE) for a transition from n to n=1 is:
\[ \Delta E = E_1 - E_n \]
For n=2:
\[ E_2 = -\frac{2.18 \times 10^{-8}}{2^2} = -\frac{2.18 \times 10^{-8}}{4} = -5.45 \times 10^{-9} \text{ J} \]
So,
\[ \Delta E = (-2.18 \times 10^{-8}) - (-5.45 \times 10^{-9}) = -1.635 \times 10^{-18} \text{ J} \]

For n=3:
\[ E_3 = -\frac{2.18 \times 10^{-8}}{3^2} = -\frac{2.18 \times 10^{-8}}{9} = -2.42 \times 10^{-9} \text{ J} \]
So,
\[ \Delta E = (-2.18 \times 10^{-8}) - (-2.42 \times 10^{-9}) = -1.936 \times 10^{-18} \text{ J} \]

### Step 4: Convert Energy to Wavelength
Using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h = 6.626 \times 10^{-34} \text{ J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \text{ m/s} \) (speed of light)

Rearranging gives:
\[ \lambda = \frac{hc}{E} \]

Calculating for the energy difference calculated for n=2:
\[ \lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{1.635 \times 10^{-18}} \approx 1.215 \times 10^{-7} \text{ m} = 121.5 \text{ nm} \]

### Step 5: Determine the Spectral Range
The Lyman series transitions from n=2 to n=1 gives us a wavelength of approximately 121.5 nm, which falls in the ultraviolet (UV) region of the electromagnetic spectrum.

### Conclusion
The spectral range of the Lyman series lies in the UV region.

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