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 $(m_p=1860m_e)$

• 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, ${\lambda }_H=1028\mathring{A},m_p=1860m_e$
$\therefore {\lambda }_D=\left(1-\frac{1}{2\times 1860}\right){\lambda }_H=\left(1-\frac{1}{3720}\right){\lambda }_H$
${\lambda }_D=0.99973$
${\lambda }_H=0.99973\times 1028\mathring{A}$
$=1027.7\mathring{A}$