Question

If wavelength of the first line of the Balmer series of hydrogen atom is $656.1nm$. Then wavelength of the second line of this series would be:

• A. $218.7nm$
• B. $328nm$
• C. $486nm$
• D. $640nm$

Answer 1

Answer: (C)

$486nm$

For Balmer series, $\frac{1}{{\lambda }_1}=R(\frac{1}{4}-\frac{1}{9})\Rightarrow \frac{1}{{\lambda }_1}=\frac{5R}{36}$

$n_1=2,n_2=3$

For $2nd$ line $n_2=4$, $\frac{1}{{\lambda }_2}=R(\frac{1}{4}-\frac{1}{16})=\frac{1}{{\lambda }_2}=\frac{3R}{16}$

Dividing $\frac{{\lambda }_2}{{\lambda }_1}=\frac{5R}{36}\times \frac{164}{3R}\Rightarrow {\lambda }_2=\frac{20}{9\times 3}\left(656.1\right)nm$

${\lambda }_2=486nm$ .