Question

The wavelength of first line of the Lymen series for hydrogen is $1216A^o$. The wavelength for the first line of this series for a $10$ time ionised sodium atom ($z=11$) will be :

• A. $1000A$
• B. $100A$
• C. $10A$
• D. $1A$

Answer 1

Answer: (C)

$10A$

Wavelength of transition in first Lymen series for $H$ or $H$-like atoms is given by :

$\lambda =\frac{1}{-R{Z}^2(1/n_2^2-1/n_1^2)}$

where, $n_1=2,n_2=1$

Given : ${\lambda }_H=\frac{1}{-R{1}^2(1/1^2-1/2^2)}=\frac{4}{3R}=1216A^{\circ }$

${\lambda }_{Na}=\frac{1}{-R{11}^2(1/1^2-1/2^2)}=\frac{4}{3\times {11}^{2}R}=1216/{11}^2=10.04A^{\circ }$

Hence the option C is correct.