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  = λmin =  ?

(b) Longest Wavelength = λmax =  ? 



Solution:

Rydberg formula for Paschen Series:

1λ = R( 

where n = 4,5,6,7,..............


(a) Shortest Wavelength = \λ_{min} =  ?

For the shortest Wavelength the number of orbitals is used:
n = ∞ (ie. the electron jumps from infinite orbital to the orbital no. 3)

Using Rydberg formula:

\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} =  ?

Now for the longest Wavelength, the number of orbital n = 4 s is used (ie. the electron jumps from orbital 4 to orbital no. 3)


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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