Question

If the wavelength of the first line of Balmer series of hydrogen is $6561\mathring{\text{A}}$, the wavelength of the second line of the same series will be,

• A. $13122\mathring{\text{A}}$
• B. $3280\mathring{\text{A}}$
• C. $4860\mathring{\text{A}}$
• D. $2187\mathring{\text{A}}$

Answer 1

The correct option is C $4860\mathring{\text{A}}$

The wavelength of the spectral line in Balmer series is given by
$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{n^2})$

For first line of Balmer series $n=3$
$\Rightarrow \frac{1}{{\lambda }_1}=R(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5R}{36}-\left(1\right)$

For second line of Balmer series $n=4$
$\Rightarrow \frac{1}{{\lambda }_2}=R(\frac{1}{2^2}-\frac{1}{4^2})=\frac{3R}{16}-\left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$, we get,
$\Rightarrow \frac{\frac{1}{{\lambda }_1}}{\frac{1}{{\lambda }_2}}=\frac{\frac{5R}{36}}{\frac{3R}{16}}=\frac{5}{3}\times \frac{{16}^4}{{36}^9}=\frac{20}{27}$

$\Rightarrow {\lambda }_2=\frac{20}{27}{\lambda }_1$

$\Rightarrow {\lambda }_2=\frac{20}{27}\times 6561=4860\mathring{\text{A}}$

Hence, $\left(C\right)$ is the correct answer.