Question

The wavelength of spectral line in the Lyman series of a $H$-atom is $1028 \mathring{A}$. If instead of hydrogen, we consider deuterium then shift in the wavelength of this line will be $(m_p = 1860 m_e)$

• A. $1027.7\,\mathring{A}$
• B. $1036\,\mathring{A}$
• C. $1028\,\mathring{A}$
• D. $1021\,\mathring{A}$

Answer 1

Answer: (A)

If $\lambda_D$ and $\lambda_H$ are wavelengths emitted in the case of deuterium and hydrogen
$\therefore \frac{\lambda_D}{\lambda_H} = (1 - \frac{m_e}{2m_p})$
Here, $\lambda_{H} = 1028 \mathring{A}, m_{p} = 1860 m_{e} $
$ \therefore \lambda_{D} = \left(1- \frac{1}{2\times1860}\right) \lambda_{H} = \left( 1- \frac{1}{3720}\right)\lambda_{H} $
$ \lambda_{D} = 0.99973$
$\lambda_{H} = 0.99973 \times 1028 \mathring{A} $
$= 1027.7 \mathring{A}$