Question

The wavelength of the first member of Balmer series of hydrogen is $6563\times {10}^{-10}\text{m}$. Calculate the wavelength of its second member.

• A. $4861\times {10}^{-10}\text{m}$
• B. $4861\times {10}^{-8}\text{m}$
• C. $5850\times {10}^{-10}\text{m}$
• D. $5850\times {10}^{-8}\text{m}$

Answer 1

The correct option is A $4861\times {10}^{-10}\text{m}$

For the first member of the Balmer series, we have
$\frac{1}{{\lambda }_1}=R(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5}{36}R...\left(i\right)$
For the second member
$\frac{1}{{\lambda }_2}=R(\frac{1}{2^2}-\frac{1}{4^2})=\frac{3}{16}R...\left(ii\right)$
Dividing (i) by (ii)
$\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\frac{20}{27}$
$\Rightarrow {\lambda }_2=\frac{20}{27}{\lambda }_1$
$\Rightarrow {\lambda }_2=\frac{20\times 6563\times {10}^{-10}}{27}=4861\times {10}^{-10}\text{m}$