Bohr atom: A hydrogen atom makes a downward transition from the state to the n = 5 state. Find the wavelength of the emitted photon. The lowest level energy state of hydrogen is -13.6 eV. (h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J, c = 3.00 × 10⁸ m/s) A) 2.43 μm B) 1.46 μm C) 1.94 μm D) 2.92 μm

Question

Quizplus · Accepted Answer

The energy difference between the initial and final state is given by:

ΔE = Ei - Ef

where Ei and Ef are the energies of the initial and final states, respectively.

Ei can be calculated using the formula for the energy of a hydrogen atom in the nth energy level:

Ei = -13.6/n^2 eV

where n is the principal quantum number of the energy level.

For the initial state, n = 11, so:

Ei = -13.6/11^2 eV = -0.109 eV

Ef is the energy of the final state, which is the n = 5 energy level. Using the same formula:

Ef = -13.6/5^2 eV = -0.544 eV

The energy difference is therefore:

ΔE = Ei - Ef = 0.435 eV

To find the wavelength of the emitted photon, we can use the formula:

λ = hc/ΔE

where h is Planck's constant, c is the speed of light, and ΔE is the energy difference in joules.

Converting the energy difference from eV to joules:

ΔE = 0.435 eV × (1.60 × 10^-19 J/eV) = 6.96 × 10^-20 J

Substituting the values:

λ = (6.626 × 10^-34 J ∙ s) × (3.00 × 10^8 m/s) / (6.96 × 10^-20 J) = 2.43 μm

Therefore, the correct answer is A) 2.43 μm.

Bohr Atom: a Hydrogen Atom Makes a Downward Transition from the