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Two step process. First this formula finding period, which is Hertz. (s -1)( Z number lithium = 3 )

Hertz = (3.29 X 1015 s -1)*(Z2)*[1/Nf2 - 1/Ni2]

Hertz = (3.29 X 1015 s -1)*(32)*[1/22 - 1/42]

Hertz = (3.29 X 1015 s -1)*(9)*(1/4 - 1/16)

= 5.55 X 1015 Hertz

------------------------------now,

Wavelength = speed of light/Hertz

Wavelength = 2.998 X 108 (m/s)/5.55 X 1015 Hertz (s -1)

= 5.40 X 10 -8 meters

================== ( 54 nanometers ????? Seems unreasonable, but the formulas and work check )

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βˆ™ 11y ago
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βˆ™ 3mo ago

The wavelength of light emitted during this transition can be calculated using the Rydberg formula: ( \frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) ), where ( n_f = 2 ) and ( n_i = 4 ) for the given transition. Lithium has an atomic number of 3, so the Rydberg constant ( R_H = 1.097 \times 10^7 m^{-1} ). Substituting these values into the formula will give you the wavelength of light emitted.

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βˆ™ 13y ago

I don't think that a chemical has a wavelength by itself. It depends on what you did that to that chemical, like if you burned it, it emits a red color. Red's wavelength is 700 nanometers.

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βˆ™ 14y ago

670 nanometers

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Q: What is the wavelength of light emitted during a transition from the 4d orbital to the 2p orbital in Lithium?
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