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The hydrogen atom, changing from its first excited state to its lowest energy state, emits light with a wavelength of 122 nm. That is in the far ultraviolet. The sodium atom, which like hydrogen has one electron that gets excited outside a core of 10 other electrons, emits light at 589 nm making a similar transition from its first excited state to its lowest state. Which of these statements would be true about the sodium and hydrogen atoms and their spectra?

a. There would be other series of spectral lines ending on the first excited states of both atoms.
b. There would be a series of spectral lines in hydrogen with the longest wavelength one at 122 nm.
c. There would be a series of spectral lines in sodium with the longest wavelength one at 589 nm.
d. The hydrogen atom binds its electron more tightly than the sodium atom does, and would require more energy to remove its electron completely.

Sagot :

Answer:

true b and c

Explanation:

n the electromechanical transitions of the atoms the relationship must be fulfilled

        [tex]\frac{x}{\lambda }[/tex] = R (1 / nf - 1 / no²)

where for the final state nf = 1 giving in the case of hydrogen the Lymma series whose smallest wavelength is lam = 122 nm with nf = 1 and there are a series of spectral lines for each value of n of the final state

in the case of sodium so well it has a transition from an excited state to the kiss state (bad)

Now let's review the different proposals

a) False. The electronic potential for sodium is much lower than for hydrognosia

b) True

c) True

d) true

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