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  • In Balmer series of hydrogen atom, the wavelength of II nd line (also called as  - line) is :-<s

In Balmer series of hydrogen atom, the wavelength of II nd line (also called as \beta - line) is :-

Option: 1

\mathrm{656.28 A^{\circ }}


Option: 2

\mathrm{4872 A^{\circ }}


Option: 3

\mathrm{6562.8 A^{\circ }}


Option: 4

\mathrm{486.2 A^{\circ }}


Answers (1)

best_answer

The transition responsible for IInd Balmer \mathrm{(\beta -line)} line is 4 \rightarrow 2. In H-atom, \mathrm{n_1=2} for Balker series.

\begin{aligned} & \mathrm{\Rightarrow \quad \Delta E=13.6(1)^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=2.55 \mathrm{eV}} \\ \\&\mathrm{ \lambda=\frac{h c}{\Delta E}=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2.55 \times 1.6 \times 10^{-19}}} \\ \\& \mathrm{\lambda=4.872 \times 10^{-7} \mathrm{~m}=4872\: A^{\circ}} . \end{aligned}

Option (2) is the correct answer.

Posted by

Gautam harsolia

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