Light of wavelength 411 nm is observed from a hydrogen discharge tube. What transition produces this emission? The energy of the n = 1 level is -13.6 eV.
A) n = 1.13 to n =1
B) n = 6 to n = 2
C) n = 3 to n = 3
D) n = 2 to n = 6
E) n = 5 to n = 1

Question

Quizplus · Accepted Answer

The wavelength of 411 nm corresponds to a transition between the energy levels of n=6 and n=2. We can use the formula for the energy of a hydrogen atom in a particular energy level to confirm this:

E = -13.6 eV/n^2

For n=2, the energy is:

E = -13.6 eV/2^2 = -3.4 eV

For n=6, the energy is:

E = -13.6 eV/6^2 = -0.75 eV

The difference in energy between these levels is:

ΔE = (-0.75 eV) - (-3.4 eV) = 2.65 eV

We can use the formula for the energy of a photon to calculate the wavelength:

E = hc/λ

where h is Planck's constant, c is the speed of light, and λ is the wavelength. Solving for λ:

λ = hc/E = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(2.65 eV)(1.60 x 10^-19 J/eV) = 410 nm

This is a very close match to the observed wavelength of 411 nm, which confirms that the transition between n=6 and n=2 is responsible for the emission.

Light of Wavelength 411 Nm Is Observed from a Hydrogen