[Coastie17]

Just need an answer check. Thank you.

Determine the energy of a photon associated with the series limit of the Humphreys (n=6) series.

v-tilde = 109680[(1/6²) + (1/∞²)]cm^-1 = 3046.7 cm^-1 (100 cm/1m) = 304670 m^-1

wavelength = 1/v-tilde = 1/304670 m^-1 = 3.028 x 10^-6 m

c/wavelength = v  (299792458 m/s)/3.28 x 10^-6 m = 9.14 x 10¹³ s^-1

E = hv = (6.626 x 10^-34 J s)(9.14 x 1013 s^-1) = 6.06 x 10^-20 J

[Borek]

Logic looks OK. 3.028x10^-6 m is wrong, but later you use a correct value of 3.28x10^-6 m, so it is probably just a typo.

Not that I checked numerical values.