The shortest wavelength in the Lyman series of the hydrogen spectrum is 912 Å. The shortest wavelength present in Paschen series of spectral lines will be:

Concept:

Bohr model:

The Lyman series: 

• It includes the lines emitted by transitions of the electron from an outer orbit of quantum number n2 > 1 to the 1st orbit of quantum number n1 = 1.
• All the energy wavelengths in the Lyman series lie in the ultraviolet band.

The Balmer series:

• It includes the lines due to transitions from an outer orbit n2 > 2 to the orbit n1 = 2.
• Four of the Balmer lines lie in the "visible" part of the spectrum.

Paschen series (Bohr series, n1 = 3)

Brackett series (n1 = 4)

Pfund series (n1 = 5)​

Paschen series: 

• When an electron in a Hydrogen atom transit from a higher energy orbit to 3rd orbit. (outer orbit n2 = n > 3 to the orbit n1 = 3) known as Paschen Series.
• So the empirical formula for the observed wavelengths (λ)for hydrogen (Z = 1) is

\(\Rightarrow {\displaystyle {1 \over λ }=R\left({1 \over {n_{1}}^{2}}-{1 \over {n_{2}}^{2}}\right)}\)

For λmin, n2 = ∞ 

Calculation:

Given: 

The shortest wavelength in the Lyman series of the hydrogen spectrum = 912 Å.

\(\Rightarrow {\displaystyle {1 \over λ_{min} }=R\left({1 \over {1}^{2}}-0\right)} = { R }\)

\({λ _{min}} = \frac{1}{R}\)  =  912 Å. -----(1)

The shortest wavelength of the Paschen series means

For λmin,  n1 = 3, n2 = ∞ 

\(\Rightarrow {\displaystyle {1 \over λ_{min}' }=R\left({1 \over {3}^{2}}-0\right)} = \frac{ R }{9}\)

\({λ _{min}}' = \)  9/R ------(2)

Putting the value of (1) in (2), we get,

\({λ _{min}}' = \) 8208 Å

The correct answer is option (1)