An electron moves from an excited state back to the ground state releasing 4.56 x 10-19 J of energy. a) Find the frequency of the photon of light that would be released. b) Find the
wavelength in nm.

Explanation:

The first step in this problem is to convert the wavelength, λ, to a frequency, ν, and

then calculate the energy of the photon using ε = hν. In such problems you must take

care to use a consistent set of units. We will use S.I. units. Thus,

λ = 18900 Å = 18900 x 10–10 m = 1.89 x 10–6 m

The frequency associated with this wavelength is given by:

ν = c/λ = (3.00 x 108 m s–1 ) / (1.89 x 10–6 m) = 1.59 x 1014 s–1

The energy ε associated with a photon of this frequency is:

ε = h ν = (6.626 x 10–34 J s) (1.59 x 1014 s–1) = 1.05 x 10–19 J

Note -- the ENERGY of this photon corresponds to a DIFFERENCE between

energy levels of the H atom

The energy levels for the H atom are given by:

εn = (–21.8 x 10–19 J) / n2 ; where n = 1,2,3,4,

Thus, ε1 = –21.8 x 10–19 J; ε2 = –5.45 x 10–19 J

ε3 = –2.42 x 10–19 J; ε4 = –1.36 x 10–19 J

We now seek an ENERGY LEVEL DIFFERENCE which matches the energy of the

photon.

We see that the electronic transition which gives rise to an EMITTED photon with

energy ε = 1.05 x 10–19 J is:

ninitial = 4 → nfinal = 3

2. A series of spectral lines in emission is characterized by a common value of nfinal -- i.e.

the quantum number of the level at which the transition terminates. Since the 18900 Å

line arises from a transition in which ninitial = 4 and nfinal = 3, the series to which it

belongs is characterized by nfinal = 3. The series limit for this series of emission lines is

the wavelength that would correspond to the transition ninitial = ∞ to nfinal = 3

From the energy level expression given in problem1 we see that:

ε∞ = (–21.8 x 10–19 J) (1/ ∞2) = 0

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.