# Five lines in the H atom spectrum have wavelengths (in nm): a) 121.27 n(initial): ____ b) 4334.05 n of initial: _____ c) 486.13 n of initial: ______ d) 656.28 n of initial: _____ e) 1093.8 n of...

Five lines in the H atom spectrum have wavelengths (in nm):

a) 121.27 n(initial): ____

b) 4334.05 n of initial: _____

c) 486.13 n of initial: ______

d) 656.28 n of initial: _____

e) 1093.8 n of initial: _____

Three lines result from transitions to n of final=2 (visible series). The other two result from transition in different series, one with n of final =1 and the other with n final=3

Identify n of initial for each line.

If you can just show how to do the first one, it will help me greatly for the others when I do it myself.

### 1 Answer | Add Yours

The emission spectrum of atomic hydrogen is divided into a number of spectral series, with wavelengths given by the Rydberg formula. These observed spectral lines are due to transition of electrons between energy levels in the atom. Rydberg formula for emission spectrum of hydrogen atom is:

`1/lambda=(1/n_f^2-1/n_i^2)`

Where R is Rydberg constant, its value in the S.I. units is `1.097*10^7m^(-1)`

Electronic transition from higher energy levels to n=1 gives rise to the Lyman series of spectral lines in the UV region (approximately 1-400 nm of wavelength) of the electromagnetic spectrum.

Checking the wavelengths of given spectral lines, it is evident that line a) is one of the Lyman series (`n_f=1` ).

Plugging in the value of lambda in corresponding Rydberg equation,

`1/(121.27*10^(-9)) = 1.097*(1-1/n_i^2)`

`rArr n_i= 2`

Similarly, lines c) d) and e) belongs t the visible region of the electromagnetic spectrum. Hence these belong t the Balmer series (`n_f=2` ).

Corresponding Rydberg equation will be

`1/lambda=(1/2^2-1/n_i^2)`

Plugging in the wavelengths, corresponding `n_i ` can be derived.

Similarly line b) can be shown to belong to the IR region (hence Paschen series) of H-atom spectrum. The `n_i` can be obtained from the corresponding Rydberg equation:

`1/lambda=(1/3^2-1/n_i^2)`

**Sources:**