Question

According to the Bohr model, the energy of the hydrogen atom is given by the equation: E = (-21.7 x 10 -19 J)/ n 2 Calculate the wavelength of the photon emitted when the atom undergoes relaxation from the first excited state to the ground state.

Answer 1

Answer:

91.6 nm

Explanation:

The energy of the hydrogen atom can be calculated by the emission of a photon. When an electron is excited it goes from to the next energetic level, and when it returns to its ground state, it emits a photon. Hydrogen has only one electron, which is at the level n = 1. So, the equation is given:

E = (-21.7x10⁻¹⁹J)/1²

E = -21.7x10⁻¹⁹J

The energy of the photon is the energy absorbed, and because of that is positive (the opposite of the energy released by the electron). This energy can be calculated by:

E = h*c/λ

Where h is the Planck's constant (6.626x10⁻³⁴ J.s), c is the speed of the light (3.00x10⁸ m/s), and λ is the wavelength of the photon.

21.7x10⁻¹⁹ = 6.626x10⁻³⁴ * 3.00x10⁸/λ

λ = 9.16x10⁻⁸ m

λ = 91.6 nm