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Electron Transitions in Hydrogen
Electrons in hydrogen atoms can gain energy and transition between energy levels.
- Example: An electron in the ground state (n = 1) can be promoted to a higher energy state (n = 3).
- When the electron falls back from n = 3 to n = 2, it emits light at a specific wavelength.Emission Spectrum of Elements
- The light emitted during these transitions results in lines in the emission spectrum of hydrogen.
- The emitted light encompasses various regions of the electromagnetic spectrum, including ultraviolet and infrared.
Focus on the Hydrogen Atom
Hydrogen is particularly studied due to its simpler mathematical treatment compared to multi-electron atoms.
- The wavelengths of the lines in hydrogen's emission spectrum are predictable by calculating energy differences between states.For an electron in energy state n, the number of possible transitions (lines) it can produce is n - 1:
- E.g., if n = 3, possible transitions = 3 - 1 = 2 lines.The Bohr model and quantum mechanical models can accurately predict the emission spectrum for hydrogen, a single-electron system.
Electromagnetic Spectrum Regions
- Emission lines viewed through a spectroscope show only three visible lines due to hydrogen's transitions:
- Ultraviolet Part: Emission when an electron transitions from n = 1 to a higher state and back to n = 1.
- Visible Part: When the electron transitions from n = 2 to higher energy states and back to n = 2.
- Infrared Part: Emission occurring when transitions from n = 3 to higher energy and back to n = 3.
Example Calculation of Wavelength
Problem Statement: Determine the wavelength of light emitted when an electron transitions from n = 7 to n = 2.
- Formulas Used:
- Energy of emitted photon:
E_{ ext{photon}} = 2.18 imes 10^{-18} ext{ joules} imes igg(\frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2}\bigg)
- Wavelength relation:
- Rearranged to:Calculations:
- Initial and final states identified:
- n_final = 2
- n_initial = 7
- Squared values:
-
-
- Energy equation becomes:
- Calculation:
- Compute the energy difference:
- Result:Finding the wavelength:
- Constants:
- Planck’s constant:
- Speed of light:
- Plugging values into wavelength formula:
- Result for wavelength:
-
- This corresponds to light emitted during the transition from n = 7 to n = 2.
Orbital Shapes and Quantum Numbers
s Orbitals (l = 0):
- Spherical shape.
- Each principal energy level (n) has one s orbital, which is the lowest energy level in its state.p Orbitals (l = 1):
- Exist in each principal energy level above n = 1, with three p orbitals.
- Shape resembles an hourglass, oriented along x, y, and z axes.d Orbitals (l = 2):
- Present in principal energy levels above n = 2, comprising five d orbitals.
- Fourth-lowest energy orbitals, complex shapes including toroidal regions.f Orbitals (l = 3):
- Exist in principal energy levels above n = 3 with seven f orbitals.
- Structures are considerably complex, some having eight lobes.
Why Atoms Are Spherical
Atoms are typically depicted as spherical due to the distribution of orbitals which, when combined, create a roughly spherical shape, akin to multiple balloons tied together.
Summary of Orbital Types:
- A one s orbital is spherical.
- p orbitals have hourglass shapes.
- d orbitals involve more complex geometries with lobes and rings.
- f orbitals present even more complicated structures with multiple lobes.
Transition to Electron Configuration
- Now moving on to Chapter Nine: Electron Configuration.
- Refer to page seven of the fourth class handout for upcoming discussion details.