Ahigh 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?

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.

to be fair, you have to have a very high iq to understand rick and morty. the humour is extremely subtle, and without a solid grasp of theoretical physics most of the jokes will go over a typical viewer's head. there's also rick's nihilistic outlook, which is deftly woven into his characterisation- his personal philosophy draws heavily from narodnaya volya literature, for instance. the fans understand this stuff; they have the intellectual capacity to truly appreciate the depths of these jokes, to realise that they're not just funny- they say something deep about life. as a consequence people who dislike rick & morty truly are idiots- of course they wouldn't appreciate, for instance, the humour in rick's existential catchphrase "wubba lubba dub dub," which itself is a cryptic reference to turgenev's russian epic fathers and sons. i'm smirking right now just imagining one of those addlepated simpletons scratching their heads in confusion as dan harmon's genius wit unfolds itself on their television screens. what fools.. how i pity them.

and yes, by the way, i do have a rick & morty tattoo. and no, you cannot see it. it's for the ladies' eyes only- and even then they have to demonstrate that they're within 5 iq points of my own (preferably lower) beforehand. nothin personnel kid