Question

Taking the wavelength of first Balmer line in hydrogen spectrum $n=3\text{to}n=2$ as $660\text{nm}$, the wavelength of the $2^{nd}$ Balmer line $n=4\text{to}n=2$ will be

• A. $642.7\text{nm}$
• B. $889.2\text{nm}$
• C. $388.9\text{nm}$
• D. $488.9\text{nm}$

Answer 1

The correct option is D $488.9\text{nm}$

Given:
${\lambda }_1=660\text{nm}$,
Wavelength of first Balmer line
$\frac{1}{{\lambda }_1}=R(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5R}{36}---\left(1\right)$
Wavelength of second Balmer line
$\frac{1}{{\lambda }_2}=R(\frac{1}{2^2}-\frac{1}{4^2})=\frac{3R}{16}---\left(2\right)$
From $\left(1\right)\text{and}\left(2\right)$
$\therefore \frac{{\lambda }_2}{{\lambda }_1}=\frac{80}{108}$

${\lambda }_2=\frac{80}{108}{\lambda }_1=\frac{80}{108}\times 660$
${\lambda }_2=488.9\text{nm}$
Hence option $\left(B\right)$ is correct.