Hydrogen spectrum

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Question

Determine the separation of the first line of the Balmer series (in $Å$ ) in a spectrum of ordinary hydrogen and tritium (mass number 3). Take Rydberg's constants = 10967800 m^-1.

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Solution

First line of Balmer series of H-atom is
$\frac{1}{λ_{H}} = R (\frac{1}{4} - \frac{1}{9}) \times \frac{m_{H}}{m_{e} + m_{H}}$
First line of Balmer series for Z-atom is
$\frac{1}{λ_{T}} = R (\frac{1}{4} - \frac{1}{9}) \times \frac{m_{T}}{m_{e} + m_{H}}$
$\Rightarrow λ_{H} - λ_{T} = \frac{36 (m_{e} + m_{H})}{5 R m_{H}} - \frac{36 (m_{e} + m_{T})}{5 R m_{T}}$
$\begin{matrix} = \frac{36}{5 R} [\frac{(m_{e} + m_{H}) m_{T} - (m_{e} + m_{T}) m_{H}}{m_{H} m_{T}}] \\ = \frac{36}{5 R} [\frac{m_{e} (m_{T} - m_{H})}{m_{H} m_{T}}] \\ = \frac{36}{5 \times 10967800} \times \frac{9.109 \times 10^{- 31} \times 2 \times 1.67 \times 10^{- 27}}{3 \times (1.67 \times 10^{- 27})} = 2.387 Å \end{matrix}$

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Hydrogen spectrum

Remember concepts with our Masterclasses.

Determine the separation of the first line of the Balmer series (in Å) in a spectrum of ordinary hydrogen and tritium (mass number 3). Take Rydberg's constants = 10967800 m-1.

First line of Balmer series of H-atom is1λH=R14−19×mHme+mHFirst line of Balmer series for Z-atom is1λT=R14−19×mTme+mH⇒λH−λT=36me+mH5RmH−36me+mT5RmT=365Rme+mHmT−me+mTmHmHmT=365RmemT−mHmHmT=365×10967800×9.109×10−31×2×1.67×10−273×1.67×10−27=2.387Å

Ready to Test Your Skills?

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Determine the separation of the first line of the Balmer series (in $Å$ ) in a spectrum of ordinary hydrogen and tritium (mass number 3). Take Rydberg's constants = 10967800 m^-1.