calculate the wave length of the second line of Lyman series in hydrogen spectra.( Given RH =1.09737)
raviprakashsingh:
is there anyone who is going to give me answer
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We have formula ,
1/λ = RZ² [ 1/ni² - 1/nf²]
Here λ is wavelength , R is Rydberg constant i.e., Rh = 109737 cm⁻¹
For Lyman series ni = 1 and nf = 2, 3, 4, ....... ∞
But question is saying about 2nd line of Lyman series
means , ni = 1 and nf = 3
So, 1/λ = 109737 cm⁻¹ × (1)² [ 1/1² - 1/3² ] [ for Hydrogen Spectra Z = 1 ]
= 109737 × [1 - 1/9 ]
= 109737 × 8/9 cm⁻¹
= 877896/9cm⁻¹
Now, λ = 9/877896 cm = 1.025 × 10⁻⁷ m = 102.5 nm
1/λ = RZ² [ 1/ni² - 1/nf²]
Here λ is wavelength , R is Rydberg constant i.e., Rh = 109737 cm⁻¹
For Lyman series ni = 1 and nf = 2, 3, 4, ....... ∞
But question is saying about 2nd line of Lyman series
means , ni = 1 and nf = 3
So, 1/λ = 109737 cm⁻¹ × (1)² [ 1/1² - 1/3² ] [ for Hydrogen Spectra Z = 1 ]
= 109737 × [1 - 1/9 ]
= 109737 × 8/9 cm⁻¹
= 877896/9cm⁻¹
Now, λ = 9/877896 cm = 1.025 × 10⁻⁷ m = 102.5 nm
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