Question

Calculate the wavelength (in nanometers) of a photon emitted by a hydrogen atom when its electron drops from the n = 4 to n = 2 state.

Consider the following energy levels of a hypothetical atom:
E₄ −1.61 × 10⁻¹⁹ J

E₃ −7.51 × 10⁻¹⁹ J

E₂ −1.35 × 10⁻¹⁸ J

E₁ −1.45 × 10⁻¹⁸ J

(a) What is the wavelength of the photon needed to excite an electron from E₁ to E₄?
____ ×10m

(b) What is the energy (in joules) a photon must have in order to excite an electron from E₂ to E₃?
___× 10J

(c) When an electron drops from the E₃ level to the E₁ level, the atom is said to undergo emission. Calculate the wavelength of the photon emitted in this process.
___×10 m

Answer 1

a) ∆E =E₄ - E₁ = −1.61 × 10⁻¹⁹ J - (−1.45 × 10⁻¹⁸ J ) = 1.289x10⁻¹⁸ J

∆E = hc/ λ = ( 6.625x10^-34Js x 3x10⁸ms^-1) / (1.289x10⁻¹⁸ J) = 15.41 x 10^-8 m = 154 nm

The photon needed to excite an electron from E₁ to E₄ is 154 nm

b) ∆E =E2 - E3 = −1.35 × 10⁻¹⁸ J - (−7.51 × 10⁻¹⁹ J ) = 0.599x10⁻¹⁸ J = 5.99x10⁻¹⁹ J

The energy (in joules) a photon must have in order to excite an electron from E2 to E3 is 5.99x10⁻¹⁹ J

c) ∆E =E1 - E3 = −1.45 × 10⁻¹⁸ J - (−7.51 × 10⁻¹⁹ J ) = 0.699x10⁻¹⁸ J

∆E = hc/ λ = (6.625x10^-34Js x 3x10⁸ms^-1) / (0.699x10⁻¹⁸ J) = 28.4 x 10^-8 m = 284 nm

The wavelength of the photon emitted in this process is 284 nm