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

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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