tirria0ox

2023-02-09

The shortest and the longest wavelength in Balmer series of hydrogen spectrum are: Rydberg constant, ${R}_{H}=109678\text{}c{m}^{-1}$

Miracle Hurley

Beginner2023-02-10Added 9 answers

The energy difference between the two states exhibiting the transition should be at its maximum for the shortest wavelength in the Balmer series ${n}_{2}=\mathrm{\infty}$

So, $1\lambda ={R}_{H}=[{12}^{2}-{\mathrm{\infty}}^{2}]={R}_{H}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =4109678=3.647\times {10}^{-5}=3647\text{}A$

So, $1\lambda ={R}_{H}=[{12}^{2}-{\mathrm{\infty}}^{2}]={R}_{H}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =4109678=3.647\times {10}^{-5}=3647\text{}A$

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