Calculate the frequency of the wavelength having a wavelength of [tex]$631 \times 10^{-8}$[/tex]. Calculate the wavelength of the first two lengths of the Balmer series. Calculate the wavelength of the Lyman series.
Answer (1)
Calculate the frequency using f = λ c , resulting in f = 4.75 × 1 0 14 Hz.
Calculate the first two Balmer series wavelengths using λ 1 = R ( 2 2 1 − n 2 1 ) for n = 3 and n = 4 , resulting in λ = 6.56 × 1 0 − 7 m and λ = 4.86 × 1 0 − 7 m.
Calculate the first two Lyman series wavelengths using λ 1 = R ( 1 2 1 − n 2 1 ) for n = 2 and n = 3 , resulting in λ = 1.22 × 1 0 − 7 m and λ = 1.03 × 1 0 − 7 m.
The frequency and the wavelengths are: f = 4.75 × 1 0 14 Hz , λ Balmer , n = 3 = 6.56 × 1 0 − 7 m , λ Balmer , n = 4 = 4.86 × 1 0 − 7 m , λ Lyman , n = 2 = 1.22 × 1 0 − 7 m , λ Lyman , n = 3 = 1.03 × 1 0 − 7 m
Explanation
Problem Overview We are given a wavelength and asked to calculate its frequency. We are also asked to calculate the wavelengths of the first two lines of the Balmer and Lyman series.
Calculating Frequency First, let's calculate the frequency ( f ) of the given wavelength ( λ = 631 × 1 0 − 9 m). We know that the speed of light ( c ) is approximately 3 × 1 0 8 m/s. The relationship between frequency, wavelength, and the speed of light is given by the formula: f = λ c Substituting the given values: f = 631 × 1 0 − 9 m 3 × 1 0 8 m/s = 4.75 × 1 0 14 Hz So, the frequency is 4.75 × 1 0 14 Hz.
Balmer Series Wavelengths Next, we need to calculate the wavelengths of the first two lines of the Balmer series. The Balmer series corresponds to electron transitions where the electron falls to the n = 2 energy level. The Rydberg formula is given by: λ 1 = R ( 2 2 1 − n 2 1 ) where R is the Rydberg constant ( R = 1.097 × 1 0 7 m − 1 ) and 2"> n > 2 .
For the first line of the Balmer series, n = 3 :
λ 1 = 1.097 × 1 0 7 ( 4 1 − 9 1 ) = 1.097 × 1 0 7 ( 36 9 − 4 ) = 1.097 × 1 0 7 ( 36 5 ) = 1.5236 × 1 0 6 λ = 1.5236 × 1 0 6 1 = 6.56 × 1 0 − 7 m For the second line of the Balmer series, n = 4 :
λ 1 = 1.097 × 1 0 7 ( 4 1 − 16 1 ) = 1.097 × 1 0 7 ( 16 4 − 1 ) = 1.097 × 1 0 7 ( 16 3 ) = 2.0569 × 1 0 6 λ = 2.0569 × 1 0 6 1 = 4.86 × 1 0 − 7 m So, the wavelengths of the first two lines of the Balmer series are 6.56 × 1 0 − 7 m and 4.86 × 1 0 − 7 m.
Lyman Series Wavelengths Finally, we need to calculate the wavelengths of the first two lines of the Lyman series. The Lyman series corresponds to electron transitions where the electron falls to the n = 1 energy level. The Rydberg formula is given by: λ 1 = R ( 1 2 1 − n 2 1 ) where R is the Rydberg constant ( R = 1.097 × 1 0 7 m − 1 ) and 1"> n > 1 .
For the first line of the Lyman series, n = 2 :
λ 1 = 1.097 × 1 0 7 ( 1 − 4 1 ) = 1.097 × 1 0 7 ( 4 3 ) = 8.2275 × 1 0 6 λ = 8.2275 × 1 0 6 1 = 1.22 × 1 0 − 7 m For the second line of the Lyman series, n = 3 :
λ 1 = 1.097 × 1 0 7 ( 1 − 9 1 ) = 1.097 × 1 0 7 ( 9 8 ) = 9.7511 × 1 0 6 λ = 9.7511 × 1 0 6 1 = 1.03 × 1 0 − 7 m So, the wavelengths of the first two lines of the Lyman series are 1.22 × 1 0 − 7 m and 1.03 × 1 0 − 7 m.
Final Answer In summary:
The frequency of the given wavelength is 4.75 × 1 0 14 Hz.
The wavelengths of the first two lines of the Balmer series are 6.56 × 1 0 − 7 m and 4.86 × 1 0 − 7 m.
The wavelengths of the first two lines of the Lyman series are 1.22 × 1 0 − 7 m and 1.03 × 1 0 − 7 m.
Examples
Understanding the frequency and wavelength of light is crucial in various fields, such as telecommunications, medicine, and astronomy. For instance, in telecommunications, different frequencies of electromagnetic waves are used to transmit information wirelessly. In medicine, X-rays, which are part of the electromagnetic spectrum, are used for imaging bones. In astronomy, analyzing the wavelengths of light emitted by stars helps us understand their composition and distance from Earth. The Balmer and Lyman series are specific sets of wavelengths emitted by hydrogen atoms, which are fundamental in understanding atomic structure and quantum mechanics.
