Compute the shortest wavelength radiation in the Balmer series? What value of n must be used?  (Ans: 364.5 nm & n = ∞)

 


Data Given:

Balmer Series:

∴ Rydberg constant = R = 1.0974 ｘ 10⁷ m⁻¹


To Find:

(a) Shortest Wavelength in Balmer Series = `\λ_{min}` =  ?

(b) Energy level = n = ?



Solution:

Rydberg formula for Balmer Series:

`\frac {1}{λ}` = R`(\ frac {1}{2²} - frac {1}{n²})` -----(1)

where n = 3,4,5,..............


To find the Shortest Wavelength `\λ_{min}`, the energy level n = ∞ is used. (It means the electron jumps from infinite orbital to orbital no. 1)

by putting values

`\frac {1}{λ_{min}}` = 1.0974 ｘ 10⁷ m⁻¹ `(\ frac {1}{2²} - frac {1}{∞²})`


`\frac {1}{λ_{min}}` = 1.0974 ｘ 10⁷ m⁻¹ `(\ frac {1}{4} - 0 )`


`\frac {1}{λ_{min}}` = `\frac {1.0974 ｘ 10⁷ m⁻¹}{4}`    


by flipping the fraction on both sides of the equation we get


`\λ_{min}` = `\ frac {4}{1.0974 ｘ 10⁷ m⁻¹}`


or


`\λ_{min}` = 3.6449 ｘ 10⁻⁷ m


or


`\λ_{min}` = 364.5 ｘ 10⁻⁹ m


`\λ_{min}` = 364.5 nm 
and the excited energy level n = ∞------------Ans.



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