Longest and shortest wavelength in paschen series
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Answered by
11
i dont know about longest wavelength but i know about shortest one:
The Paschen’s series is given by
1/λ = R (1/(n1)2 - 1/(n2)2 ) when n1 = 3, and n2 = 4, 5, 6…
Now the wavelength will be shortest if n2 = ∞
1/λ = R (1/(3)2 - 1/(∞)2 )
=> 1/λ = 1.0974×107(1/9 – 0)
=> 1/λ = 1.0974×107/9 = 1219333 m-1
=> λ = 8201 A
hope it helps!!☺☺
The Paschen’s series is given by
1/λ = R (1/(n1)2 - 1/(n2)2 ) when n1 = 3, and n2 = 4, 5, 6…
Now the wavelength will be shortest if n2 = ∞
1/λ = R (1/(3)2 - 1/(∞)2 )
=> 1/λ = 1.0974×107(1/9 – 0)
=> 1/λ = 1.0974×107/9 = 1219333 m-1
=> λ = 8201 A
hope it helps!!☺☺
Anonymous:
the 2 after the bracket is to the power of tht bracket ....it is not fr multiplication....
Answered by
5
This is the answer for your given question please verify the final values!
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