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The first four spectral lines in the Lyman series of a H-atom are \lambda = 1218 Å, 1028Å, 974.3 Å and 951.4Å. If instead of hydrogen, we consider deuterium, calculate the shift in the wavelength of these lines.

Answers (1)

Let us take \mu H as the reduced masses of electron of hydrogen and \mu D as the reduced masses of electron of deuterium.

 we know that

\frac{1}{\lambda}=R\left [ \frac{1}{n_{f}^{2}} -\frac{1}{n_{i}^{2}}\right ]

As ni and nf are fixed for by mass series for hydrogen and deuterium

\lambda \propto \frac{1}{R}or\; \frac{\lambda_{D}}{\lambda_{H}}=\frac{R_{H}}{R_{D}}........................(i)\\ R_{R}=\frac{m_{e}e^{4}}{8\varepsilon _{0}ch^{3}}=\frac{\mu_{H}e^{4}}{8\varepsilon _{0}ch^{3}}\\ R_{D}=\frac{m_{e}e^{4}}{8\varepsilon _{0}ch^{3}}=\frac{\mu_{D}e^{4}}{8\varepsilon _{0}ch^{3}}\\ \therefore \frac{R_{H}}{R_{D}}=\frac{\mu_{H}}{\mu_{D}}.................(ii)\\

From equation (i) and (ii)

\frac{\lambda_{H}}{\lambda_{D}}=\frac{\mu_{H}}{\mu_{D}}.................(iii)\\

Reduced mass for hydrogen,

\mu_{H}=\frac{m_{e}}{1+m_{e}1M}\simeq m_{e}\left ( 1-\frac{m_{e}}{M} \right )

Reduced mass for deuterium,

\mu_{D}=\frac{2M.m_{e}}{2M\left (1+\frac{m_{e}}{2M} \right )}\simeq m_{e}\left ( 1-\frac{m_{e}}{2M} \right )

where M is mass of the proton

\frac{\mu_{H}}{\mu_{D}}=\frac{m_{e}\left (1- \frac{m_{e}}{2M} \right )}{m_{e}\left (1- \frac{m_{e}}{2M} \right )}=\left ( 1-\frac{m_{e}}{M} \right )\left ( 1-\frac{m_{e}}{2M} \right )^{-1}\\ =\left ( 1-\frac{m_{e}}{M} \right )\left ( 1+\frac{m_{e}}{2M} \right )\\ \Rightarrow \frac{\mu_{H}}{\mu_{D}}=\left ( 1-\frac{m_{e}}{2M} \right )\\ or \;\;\frac{\mu_{H}}{\mu_{D}}=\left ( 1-\frac{1}{2 \times 1840} \right )=0.99973\lambda_{H}...........(iv)\\ \left ( \because M=1840 m_{e} \right )

From (iii) and (iv)

\frac{\lambda_{D}}{\lambda_{H}}=0.99973, \;\; \lambda_{D}=0.99973 \lambda_{H}\\\\ using \; \lambda_{H}=1218 \AA ,1028\AA ,974.3\AA \; and\;951.4\AA \; we\;get \\\\ \lambda_{D}=1217.7\AA ,1027\AA ,974.04\AA ,951.1\AA \\\\ Shift\;in\;wavelength\left ( \lambda_{H}-\lambda_{D} \right )\approx \approx 0.3\AA

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