Question

1) The last two visible line observed in the Balmer series are of the wavelength 434.0 nm and 410.2 nm. For each of the wavelength,

a) Determine the frequency associated with the photon associated with this wavelength.

b) Determine the amount of energy released by this photon.

c) The Balmer series are the results of excited electron at higher level relaxing to the electronic level of principle quantum number of nf = 2. Determine the excited electronic level, ni, of this electron.

d) If this excited electron has the same wavelength as the emitted photon, determine its speed during the transition.

Answer 1

a) Determine the frequency associated with the photon of wavelength 434.0 nm and 410.2 nm.

λ = c/v Here λ is the wavelength

λ = 434 nm. = 4.34 x 10^-7m

c(speed of light) = 2.998 x 10⁸ m/s

v = c/ λ

v = 2.998 x 10^8 m/s/ 4.34 x 10^-7m

v = 6.91 x 10^14/s

λ = 410.2 nm. = 4.102 x 10^-7m

v = c/ λ

v = 2.998 x 10^8 m/s/ 4.102 x 10^-7m

v = 7.31 x 10^14/s

b) Determine the amount of energy released by this photon.

h = Plank constant = 6.626 x 10^-34 J.s

E = hv = (6.626 x 10^-34 J.s)(6.91 x 10^14/s)

E = 4.5785 x 10 ^-19 J

E = hv = (6.626 x 10^-34 J.s)(7.31 x 10^14/s)

E = 4.8436 x10 ^-19 J c)

c)

ΔE atom = ΔE photon = hv = hc/ λ = (6.626 x 10-34 Js)( 2.998 x 108 m/s)/ 4.34 x 10-7m = -4.58 x 10-19 J

ΔE is negative number because it is an emission process.

ΔE = RH [1/ni2 - 1/nf2] = -4.58 x 10-19J

= (2.179 x 10-18J) [1/ni2 - 1/nf2]

-4.58 x 10-19J/2.179 x 10-18J 1/ni2-0.2501

-0.210 + 0.250 = 0.040

ni = 1 / √ 0.040

ni = 5