Calculate the shortest and longest wavelength of radiation for the Paschen Series. (Answer: 1875 nm, 820 nm)
Data Given:
Paschen Series:
∴ Rydberg constant = R = 1.097 x 10⁷ m⁻¹
To Find:
(a) Shortest Wavelength = `\λ_{min}` = ?
(b) Longest Wavelength = `\λ_{max}` = ?
Solution:
Rydberg formula for Paschen Series:
where n = 4,5,6,7,..............
For the shortest Wavelength the number of orbitals is used:
n = ∞ (ie. the electron jumps from infinite orbital to the orbital no. 3)
n = ∞ (ie. the electron jumps from infinite orbital to the orbital no. 3)
`\frac {1}{λ_{min}}` = R`(\ frac {1}{3²} - frac {1}{n²})`
`\frac {1}{λ_{min}}` = 1.097 x 10⁷ m⁻¹ `(\ frac {1}{3²} - frac {1}{∞²})`
`\frac {1}{λ_{min}}` = 1.097 x 10⁷ m⁻¹ `(\ frac {1}{9} - 0 )`
`\frac {1}{λ_{min}}` = `\ frac {1.097 x 10⁷ m⁻¹}{9}`
by flipping the fraction on both sides of the equation we get
`\λ_{min}` = `\ frac {9}{1.097 x 10⁷ m⁻¹}`
or
`\λ_{min}` = 8.204 x 10⁻⁷ m
or
`\λ_{min}` = 820.4 x 10⁻⁹ m
`\λ_{min}` = 820.4 nm ------------Ans. 1
Or (expressing in Angstrom)
`\λ_{min}` = 8204.0 x 10¹⁰ m
`\λ_{min}` = 8204 A°
(b) Longest Wavelength = `\λ_{max}` = ?
Using the Rydberg formula for longest wavelength:
`\frac {1}{λ_{max}}` = R`(\ frac {1}{3²} - frac {1}{n²})`
by putting values
`\frac {1}{λ_{max}}` = 1.097 x 10⁻⁷ m⁻¹ `(\ frac {1}{3²} - frac {1}{4²})`
`\frac {1}{λ_{max}}` = 1.097 x 10⁻⁷ m⁻¹ `(\ frac {1}{9} - frac {1}{16})`
`\frac {1}{λ_{max}}`= 1.097 x 10⁻⁷ m⁻¹ `(\ frac {16 - 9}{144})`
`\frac {1}{λ_{max}}` = 1.097 x 10⁻⁷ m⁻¹ `(\ frac {7}{144})`
`\frac {1}{λ_{max}}` = `(\ frac {7.679 x 10⁻⁷ m⁻¹}{144})`
by flipping the fraction on both sides of the equation we get
`\λ_{max}` = `\ frac {144}{7.679 x 10⁻⁷ m⁻¹}`
or
`\λ_{max}` = 18.752 x 10⁷ m
or
`\λ_{max}` = 1875.2 x 10⁹ m
`\λ_{max}` = 1875 nm ---------------Ans. 2
Or (expressing in Angstrom)
`\λ_{min}` = 18752 x 10¹⁰ m
`\λ_{min}` = 18752 A°
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