Question

The wavelength of the second line of Balmer series in the hydrogen spectrum is $4861\mathring{A}$. The wavelength of the first line is

• A. $\frac{27}{20}\times 4861\mathring{A}$
• B. $\frac{20}{27}\times 4861\mathring{A}$
• C. $20\times 4861\mathring{A}$
• D. $4861\mathring{A}$

Answer 1

Answer: (A)

For the first line in balmer series:

$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5R}{36}$

For second balmer line:

$\frac{1}{4861}=R(\frac{1}{2^2}-\frac{1}{4^2})=\frac{3R}{16}$

Divide both equations:

$\frac{\lambda }{4861}=\frac{3R}{16}\times \frac{36}{5R}$

$\lambda =4861\times \frac{27}{20}$