Question

If the wavelength of 1st line of Balmer series of hydrogen is 6561 ?, the wavelength of the 2 line of series will be

• A. 9780 ?
• B. 4860 ?
• C. 8857 ?
• D. 4429 ?

Answer 1

Answer: (B)

4860 ?

Answer 2

For the first line of Balmer series,
$\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5R}{36}$
For the second line of Balmer series.
$\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3R}{16}$
$\therefore\frac{\lambda_{2}}{\lambda_{1}}=\frac{5R/36}{3R/16}=\frac{20}{27};$
or $\lambda_{2}=\frac{20}{27}\left(6561 ?\right)=4860 ?$