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The energy of stable states of the hydro...

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?

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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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