The energy levels of an electron in a hydrogen atom are given by the formula:
Eₙ = -13.6 eV / (n²),
where Eₙ is the energy level (in electron volts) and n is the principal quantum number.
Calculate the energy of an electron in the n=3 energy level.
Calculate the wavelength (in nm) of a photon emitted when an electron transitions from the n=4 energy level to the n=2 energy level.
Solution:
Eₙ = -13.6 eV / (n²) E₃ = -13.6 eV / (3²) E₃ = -13.6 eV / 9 E₃ ≈ -1.51 eV
Therefore, the energy of an electron in the n=3 energy level is approximately -1.51 eV.
ΔE = E₂ - E₄ = hf
Where ΔE is the change in energy, E₂ and E₄ are the energy levels, h is Planck's constant (6.63 × 10⁻³⁴ J⋅s), and f is the frequency of the emitted photon.
To find the wavelength (λ), we can use the equation:
λ = c / f
Where λ is the wavelength, c is the speed of light (3.00 × 10⁸ m/s), and f is the frequency.
First, calculate the change in energy (ΔE) using the energy equation from earlier:
ΔE = E₂ - E₄ ΔE = (-13.6 eV / (2²)) - (-13.6 eV / (4²)) ΔE = -3.4 eV + (-0.85 eV) ΔE = -4.25 eV
To convert the energy from electron volts into joules, use the conversion factor:
1 eV = 1.60 × 10⁻¹⁹ J
ΔE (in J) = -4.25 eV × (1.60 × 10⁻¹⁹ J/eV) ΔE ≈ -6.80 × 10⁻¹⁹ J
Now, find the frequency of the emitted photon (f) using the equation ΔE = hf and rearrange the equation to solve for f:
f = ΔE / h f = (-6.80 × 10⁻¹⁹ J) / (6.63 × 10⁻³⁴ J⋅s) f ≈ -1.026 × 10¹⁵ Hz
Finally, calculate the wavelength (λ) using the equation λ = c / f:
λ = c / f λ = (3.00 × 10⁸ m/s) / (-1.026 × 10¹⁵ Hz) λ ≈ 2.92 × 10⁻⁷ m
Convert the wavelength from meters to nanometers:
λ (in nm) = 2.92 × 10⁻⁷ m × (1 × 10⁹ nm/m) λ ≈ 292 nm
Therefore, the wavelength of the photon emitted when the electron transitions from the n=4 energy level to the n=2 energy level is approximately 292 nm.