Asked by IRA
The lines in the visible portion of the hydrogen spectrum are called the Balmer series, for which m=2 in the Rydberg equation. Calculate the wavelength in nanometers of the spectral line in this series for which n=4.
Answers
Answered by
DrBob222
delta E = 2.180E-18 J(1/a^2 - 1/b^2) where a = 2 and b= 4) so
dE = 2.180E-18 J (1/2^2 - 1/4^2)
Solve for dE. Then dE = hc/wavelength. h is Planck's constant, c is speed of light. Solve for wavelength (in meters) and convert to nanometers. Post your work if you get stuck.
dE = 2.180E-18 J (1/2^2 - 1/4^2)
Solve for dE. Then dE = hc/wavelength. h is Planck's constant, c is speed of light. Solve for wavelength (in meters) and convert to nanometers. Post your work if you get stuck.
Answered by
IRA
So for dE I got 4.0875 × 10-19 joules
Over
(6.62607004 × 10-34 m2 kg / s)(299 792 458 m / s)
Do I have to then covert into nm?
Over
(6.62607004 × 10-34 m2 kg / s)(299 792 458 m / s)
Do I have to then covert into nm?
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