Question

a) Convert 2.08 kHz frequency into wavelength, expressing the result in m?

b) How many sets of quantum number values are there for a 2p electron?

c) One hydrogen emission line has a wavelength of 486 nm. Identify n_final and n_initial for the transition giving rise to this line?

Can you please show me all the steps. Thx

Answer 1

a)

recall that

WL = c/v

WL = wavleength, c = speed of light, v, frequnecy

so

Wl = (3*10^8) / (2.08*10^3) = 144230.769 m

b)

sets of numbers for 2p electron

n = 2, l = 1

so

ml = 0,+1,-1

ms = +1/2

i.e 3x2 = 6 positions

c)

WL = 486 nm = 486*10^-9 m

nf and ni

assume ni = 2, snce it is hydrogen

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

E = (6.626*10^-34)(3*10^8)/(486*10^-9) = 4.090123*10^-19 J/photon

now,

Apply Rydberg Formula

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)

4.090123*10^-19 = 2.178*10^-18 ( 1/nf^2 -1/(2^2))

1/nf^2 =-(4.090123*10^-19 ) / (2.178*10^-18)+ 1/4 =

nf = 4 (nearest answer)

so

4 --> drops to --> 2