Atomic hydrogen produces a well-known series of spectral lines in several regions of the electromagnetic spectrum. each series fits the rydberg equation with its own particular n1 value. calculate the value of n1 that would produce a series of lines in which the highest energy line has a wavelength of 365 nm.

n₁ = 2

Explanation:

The Rydberg formula is given by

1/λ = Rh x (1/n₁² - 1/n₂² ) where Rh = Rydberg´s constant , and n₁ and n₂ are the energy levels of the transitions involved.

The highest energy line in a series  will happen when n2 tends to infinity, and then in

1/λ = Rh x (1/n₁² - 1/n₂² )  the term 1/n₂² in the limit is zero

1/λ = Rh x 1/n₁²

So lets solve for n₁²  :

n₁²  = Rh x λ = 109,737 cm⁻¹ x 365 x 10 ⁻⁷ cm = 4.00

n₁ = √4.00 = 2

This the value for the Balmer series.

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

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