Question

The wavelength of the first line in balmer series in the hydrogen spectrum is $\lambda$. What is the wavelength of the second line :

• A. $\frac{20\lambda }{27}$
• B. $\frac{3\lambda }{16}$
• C. $\frac{5\lambda }{36}$
• D. $\frac{3\lambda }{4}$

Answer 1

The correct option is A $\frac{20\lambda }{27}$

For Balmer series in hydrogen spectrum, wavelength is given by,

$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{n^2}),n=3,4$

Now, ATQ, when n=3, $\lambda =$wavelength

$\Rightarrow \frac{1}{{\lambda }_1}=R(\frac{1}{2^2}-\frac{1}{3^2})$

${\lambda }_1$=wavelength of first line

$\Rightarrow \frac{1}{{\lambda }_1}=\frac{5R}{36}$

$R=\frac{36}{5\lambda }$ (for the first line n=3)

For the second line n=4,

$\Rightarrow \frac{1}{{\lambda }_2}=R(\frac{1}{2^2}-\frac{1}{4^2})\frac{1}{{\lambda }_2}=\frac{36}{5{\lambda }_1}\left(\frac{3}{16}\right){\lambda }_2=\left(\frac{(5\times 16){\lambda }_1}{36\times 3}\right)=\frac{20{\lambda }_1}{27}$