Question

The wavelength of spectral line in the Lyman series of a H-atom is 1028 Å. If instead of hydrogen, we consider deuterium then shift in the wavelength of this line will be $\left( m_{p} = 1860m_{e} \right)$

• A. 1027.7 Å
• B. 1036 Å
• C. 1028 Å
• D. 1021 Å

Answer 1

Answer: (A)

1027.7 Å

Answer 2

If $\lambda_{D}$ and $\lambda_{H}$ are wavelength emitted in the case of deuterium and hydrogen.

$\therefore\frac{\lambda_{D}}{\lambda_{H}} = \left( 1 - \frac{m_{e}}{2m_{p}} \right)$

Here (;gk), $\lambda_{H} = 1028Å,m_{p} = 1860m_{e}$

$\therefore\lambda_{D} = \left( 1 - \frac{1}{2 \times 1860} \right)\lambda_{H} = \left( 1 - \frac{1}{3720} \right)\lambda_{H}$

$\lambda_{D} = 0.99973\lambda_{H} = 0.99973 \times 1028Å = 1027.7Å$