Question

If the wavelength of the first line of Balmer series of hydrogen is $6561\mathring{A}$. The wavelength of the second line of the series should be-

• A. $1312\mathring{A}$
• B. $3280\mathring{A}$
• C. $4860\mathring{A}$
• D. $2187\mathring{A}$

Answer 1

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

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, $n=4$

$\Rightarrow \frac{1}{{\lambda }_2}=R(\frac{1}{2^2}-\frac{1}{4^2})=\frac{3R}{16}......\left(1\right)$

From (1) and (2) we get,

$\Rightarrow \frac{\frac{1}{{\lambda }_1}}{\frac{1}{{\lambda }_2}}=\frac{\frac{5R}{16}}{\frac{3R}{16}}=\frac{20}{27}$

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

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

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.