Lines in the Paschen series of the hydrogen spectrum are caused by the emission

Question

Lines in the Paschen series of the hydrogen spectrum are caused by the emission of accompanying the fall of an electron from outer levels to the third energy level (shell). The lines can be calculated using the Rydberg equation (listed below) where n_f = 3, n = an integer greater than 3, and R = 2.18 times 10^-18 J. Show how to determine (and then calculate) the (in k J/mole) and the corresponding wavelengths (in nm) of the first two lines in the Paschen series and indicate the region of the electromagnetic spectrum where these lines would be observed. Delta E = R (1/n^2_i - 1/n^2_f)

Explanation / Answer

first two lines:

n = 4, n = 5

a)

E = R*(1/nf^2 – 1/ni ^2)

R = -2.178*10^-18 J

Nf = final stage/level

Ni = initial stage/level

E = Energy per unit (i.e. J/photon)

E = (-2.178*10^-18)*(1/3^2 – 1/4 ^2)

E = 1.0587*10^-19

For the wavelength:

WL = h c / E

h = Planck Constant = 6.626*10^-34 J s

c = speed of particle (i.e. light) = 3*10^8 m/s

E = energy per particle J/photon

WL = (6.626*10^-34)(3*10^8)/(1.0587*10^-19)

WL = 0.00000187758 m

to nanometers:

WL = (0.00000187758)(10^9) = 1877 nm

b)

E = R*(1/nf^2 – 1/ni ^2)

R = -2.178*10^-18 J

Nf = final stage/level

Ni = initial stage/level

E = Energy per unit (i.e. J/photon)

E = (-2.178*10^-18)*(1/3^2 – 1/5 ^2)

E = 1.5488*10^-19

For the wavelength:

WL = h c / E

h = Planck Constant = 6.626*10^-34 J s

c = speed of particle (i.e. light) = 3*10^8 m/s

E = energy per particle J/photon

WL = (6.626*10^-34)(3*10^8)/(1.5488*10^-19)

WL = 0.00000128344  m

to nanometers:

WL = (0.00000128344)(10^9) = 1283 nm

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