What is the difference in energy between the two levels responsible for the ultraviolet emission line of the?
Posted on January 18th, 2010 by admin
ultraviolet emission line of the magnesium atom at 285.2nm
Please walk me through how you got the answer! 
The energy difference between the 2 energy levels involved in the UV light emission of an atom or
molecule is:
Del(E) = h x f = (h x c) / l
f would be the frequency in sec-1 (Hz), if given, but it can be calculated from f = (c / l)
where h is Planck’s constant = 6.62 x 10E-34 J.secs;
c is the speed of light = 3 x 10E08 m/s;
and l is the wavelength of the emitted light, in meters = 285.2 x 10E-09 m
Del(E) = (6.62 x 10E-34 x 3.0 x 10E10) / (2.85 x 10E-07) ~ 6.97 x 10E-17 Joules
January 18th, 2010 at 1:44 pm
The energy difference between the 2 energy levels involved in the UV light emission of an atom or
molecule is:
Del(E) = h x f = (h x c) / l
f would be the frequency in sec-1 (Hz), if given, but it can be calculated from f = (c / l)
where h is Planck’s constant = 6.62 x 10E-34 J.secs;
c is the speed of light = 3 x 10E08 m/s;
and l is the wavelength of the emitted light, in meters = 285.2 x 10E-09 m
Del(E) = (6.62 x 10E-34 x 3.0 x 10E10) / (2.85 x 10E-07) ~ 6.97 x 10E-17 Joules
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