Question

A high school student acquired an emission spectrum of an unknown sample. He knew the unknown sample contained one of the elements, whose spectra are show below. The emission spectrum of his sample showed a strong emission at 610 nm but not at 480 nm.
a. Which element did his unknown sample contain?
b.Element A above has a strong emission around 670nm. Does this emission line represent a lower energy or higher energy transition than the emission line at 428nm?

Answer 1

a. The unknown sample contained the element C.

An electromagnetic spectrum shows the electromagnetic radiation that emits (emission spectrum) or absorbs (absorption spectrum) a substance. Each spectrum is unique so its radiation can be used to identify a substance in a manner analogous to a fingerprint. Thus, when observing the spectra given for elements A, B and C, we observe that spectrum for element C is the only one that has an emission band at 610 nm but not at 480 nm, so the unknown sample corresponds to element C.

b. The 670 nm line represents a lower transition energy than an emission line at 428 nm.

The transition energy is the energy required for an electron in the sample to pass from one quantum state to another. In an emission spectrum energy is supplied to the molecules to excite them, favoring that electronic transition and measuring the energy in the form of light they emit when they relax from the quantum state to which they were excited to the fundamental or basal state in which they were initially.

This transition energy is related to the wavelength of the radiation emitted according to the following equation, which corresponds to the equation for the energy of a photon,

E = hc / λ

where E is the transition energy or the energy of the emitted photons, h is the Planck constant (6,626x10-34 J s), c is the speed of light in vacuum (3x10 8 J / s) and λ the photon's wavelength.

According to the given equation, we see that the energy of the photons changes inversely proportional to the wavelength λ. In this way,

607nm > 428 nm

Then for the transition energies corresponding to these wavelengths,

E (670 nm) < E (428nm)

So 670 nm line represents a lower transition energy than an emission line at 428 nm.