Question

The wavelength of second Balmer line in Hydrogen spectrum is $600nm$. The wavelength for its third line in Lyman series is :

• A. $800nm$
• B. $600nm$
• C. $400nm$
• D. $200nm$
• E. None of the above

Answer 1

The correct option is E None of the above

Second Balmer line is produced by transition $4\to 2$.

Hence $\frac{1}{{\lambda }_1}=K(\frac{1}{2^2}-\frac{1}{4^2})$

where ${\lambda }_1=600nm$ (Given)

Third Lyman line is produced by transition $4\to 1$.

Hence $\frac{1}{{\lambda }_2}=K(\frac{1}{1^2}-\frac{1}{4^2})$

$⟹\frac{{\lambda }_2}{{\lambda }_1}=\frac{3}{15}=1/5$

$⟹{\lambda }_2=1/5\times {\lambda }_1=1/5\times 600nm=120nm$