Question

The Hydrogen Spectrum

When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the atoms will emit electromagnetic radiation and visible light can be observed. If this light passes through a diffraction grating, the resulting spectrum appears as a pattern of four isolated, sharp parallel lines, called spectral lines. Each spectral line corresponds to one specific wavelength that is present in the light emitted by the source. Such a discrete spectrum is referred to as a line spectrum
By the early 19th century, it was found that discrete spectra were produced by every chemical element in its gaseous state. Even though these spectra were found to share the common feature of appearing as a set of isolated lines, it was observed that each element produces its own unique pattern of lines. This indicated that the light emitted by each element contains a specific set of wavelengths that is characteristic of that element.
The first quantitative description of the hydrogen spectrum was given by Johann Balmer, a Swiss school teacher, in 1885, By trial and error, he found that the correct wavelength λ of each line observed in the hydrogen spectrum was given by
1/λ = R ( 1/2^2-1/n^2)
where R is a constant, later called the Rydberg constant, and n may have the integer values 3, 4, 5, If λ is in meters, the numerical value of the Rydberg constant (determined from measurements of wavelengths) s R= 1.097x10^7 m-1
Balmer knew only the four lines in the visible spectrum of hydrogen. Thus, the original formula was written for a limited set of values of T. However, as more techniques to detect other regions of the spectrum were developed, it became clear that Balmer's formula was valid for all values of n. The entire series of spectral lines predicted by Balmers formula is now referred to as the Balmer series
Part A) What is the wavelength of the line corresponding to n =4 in the Balmer series? Express your answer in nanometers to three significant figures .
Part B) What is the wavelength of the line corresponding to n =5 in the Balmer series?

Answer 1

a) λ = 4.862 10⁻⁷ m, b) λ = 4.341 10⁻⁷ m

Step-by-step explanation:

The spectrum of hydrogen can be described by the expression

`(1)/(\lambda) = R_H ( (1)/(n_o^2) - (1)/(n^2) ) \ \ \ \ n>n_o`

in the case of the initial state n = 2 this series is the Balmer series

a) Find the wavelength for n = 4

let's calculate

`(1)/( \lambda)` = 1,097 10⁷ (
`(1)/(2^2) - (1)/(4^2)`)

\frac{1}{ \lambda} = 1.097 10⁷ 0.1875 = 0.2056 10⁷

λ = 4.862 10⁻⁷ m

b) n = 5

\frac{1}{ \lambda} = 1,097 10⁷ (
`(1)/(2^2) - (1)/(5^2)`)

\frac{1}{ \lambda} = 1.097 10⁷ 0.21 = 0.23037 10⁷

λ = 4.341 10⁻⁷ m