Question

The wavelength of first line of the Lyman series for hydrogen is $1216\stackrel{o}{A}$. The wavelength for the first line of this series for a 10 times ionised sodium atom $(Z=11)$ will be:

• A. $1216\stackrel{o}{A}$
• B. $12.16\stackrel{o}{A}$
• C. $10\stackrel{o}{A}$
• D. $110\stackrel{o}{A}$

Answer 1

Answer: (C)

$10\stackrel{o}{A}$

Wavelength is given by $\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
For the first line of Lyman series, $n_1=1$ and $n_2=2$
$\frac{1}{\lambda }=R{Z}^2[\frac{1}{1^2}-\frac{1}{2^2}]$
$\frac{1}{\lambda }=R{Z}^2\times \frac{3}{4}$
$\lambda$ is directly proportinal to $\frac{1}{Z^2}$
$\frac{{\lambda }_{Na}}{{\lambda }_H}=\frac{Z_H^2}{Z_{Na}^2}=\frac{1}{{11}^2}=\frac{1}{121}$
$\frac{{\lambda }_{Na}}{1216}=\frac{1}{121}$
$\Rightarrow {\lambda }_{Na}=\frac{1216}{121}=10A^o$