Question

An electron in He+ undergoes a transition from n = 4 to n = 2 state. Calculate the (a) energy in J, (b) frequency in Hz, (c) wavelength in nm and (d) wave number in cm-1 of the photon emitted.

Answer #1

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)

E = (-2.178*10^-18)*(1/2^2 – 1/4 ^2)

**E = 4.0837*10^-19 J/photon**

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) /(4.0837*10^-19) = 4.8676*10^-7 m

f = c/WL = (3*10^8)/(4.8676*10^-7 ) = 6.16320*10^14 Hz

c)

since WL =4.8676*10^-7 m

WL = (4.8676*10^-7) *10^9 = 486.7 nm

d)

wavenumber = 7

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