The ratio of minimum wavelengths of Lyman and Balmer series will be _______

CONCEPT:

• When atoms are excited they emit light of certain wavelengths that correspond to different colors Due to the electron making transitions between two energy levels in an atom.
• The light emission can be seen as a series of colored lines, known as atomic spectra.
  • These atomic spectra are divided into a number of spectral series. The wavelength related to these series is given by the Rydberg formula.
  • The energy differences between levels in the Bohr model is given by the Rydberg formula. Hence the wavelengths of emitted/absorbed photons

\({\displaystyle {1 \over λ }=Z^{2}R_{∞ }\left({1 \over {n_{1}}^{2}}-{1 \over {n_{2}}^{2}}\right)}\)

In the Bohr model, spectrum series are given below:

• 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.
• For minimum wavelength in any series, n₂ = ∞

CALCULATION:

• For minimum wavelength in any series, n₂ = ∞

\({\displaystyle {1 \over λ }=Z^{2}R_{∞ }\left({1 \over {n_{1}}^{2}}-{1 \over {n_{2}}^{2}}\right)}\)

\( {1 \over λ }=Z^{2}R_{∞ }{1 \over {n_{1}}^{2}}\)

\( {1 \over λ }=Z^{2}R_{∞ }{1 \over {n_{1}}^{2}}\)

for Lyman series n₁ = 1

 \( {1 \over λ_{Lyman} }=Z^{2}R_{∞ }{1 \over {1}^{2}}=Z^{2}R_{∞ }\)

for Balmer series n1 = 2

\( {1 \over λ_{Balmer} }=Z^{2}R_{∞ }{1 \over {2}^{2}}=Z^{2}R_{∞ }{1 \over 4}\)

\(\frac{1 \over λ_{Balmer}} {1 \over λ_{Lyman} }=\frac{1}{4}\)

\(\frac{λ_{Lyman} } {λ_{Balmer} }=\frac{1}{4}=0.25\)

So the correct answer is option 2.