Question

A series of lines in the spectrum of atomic H lies at wavelengths 656.46 , 486.27 , 434.17 , 410.29 nm . what is the wavelegth of the next line in this series ?

Answer 1

1/x = R*Z²*[1/(n1)²-1/(n2)²]                  where x is the wavelength

The given series lies in Balmer series as the range of wavelengths of visible spectrum is 390nm-700nm.

Hence, n1=2

Now, Case 1 , 1 / 656.46 = R*[1/(2)²-1/(n2)²]

We get n2 = 3(approx.)

Case 2 , 1 / 486.27 = R*[1/(2)²-1/(n2)²]

We get n2 = 4(approx.)

Case 3 , 1 / 434.17 = R*[1/(2)²-1/(n2)²]

We get n2 = 5(approx.)

Case 5 , 1 / 410.29 = R*[1/(2)²-1/(n2)²]

We get n2 = 6(approx.)

So, the next transition is from n2 = 7 to n1 = 2,

1/x = R*[1/(2)²-1/(7)²]  

x = 395.96 nm(approx.)