Question

The wavelength of first line of Lyman series in hydrogen atom is $1216$ $A^{\circ }$. The wavelength of first line of Lyman series for 10 times ionised sodium atom will be

• A. $0.1A^{\circ }$
• B. $1000A^{\circ }$
• C. $100A^{\circ }$
• D. $10A^{\circ }$

Answer 1

Answer: (D)

$10A^{\circ }$

For Lyman series $n_1=1$ and $n_2=2,3,...........,\infty$
Hence, the wavelength for hydrogen atom is given by the formula
$\frac{1}{{\lambda }_H}=R{Z}_H^2[\frac{1}{1^2}-\frac{1}{n_2^2}]$
$\frac{1}{1216}=R[\frac{1}{1^2}-\frac{1}{n_2^2}]$ ...................(since, Z = 1)
$\frac{1}{{\lambda }_{Na}}=R{Z}_{Na}^2[1-\frac{1}{n_2^2}]$
$\frac{1}{{\lambda }_{Na}}=R\left(11{)}^2\right[\frac{1}{1^2}-\frac{1}{n_2^2}]$

Therefore,
$\frac{\frac{1}{{\lambda }_{Na}}}{\frac{1}{1216}}=\frac{R\left(11{)}^2\right[\frac{1}{1^2}-\frac{1}{n_2^2}]}{R[\frac{1}{1^2}-\frac{1}{n_2^2}]}$

${\lambda }_{Na}=\frac{1216}{(11{)}^2}$

${\lambda }_{Na}=\frac{1216}{121}$

${\lambda }_{Na}=10.04\approx 10A^{\circ }$