Question

If the wavelength of first line of the Balmer series of hydrogen atom is $656.1$ nm, the wavelength of second line of this series would be

• A. $218.7nm$
• B. $328.0nm$
• C. $486.0nm$
• D. $640.0nm$

Answer 1

Answer: (C)

$\frac{1}{{\lambda }_1}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R_H\left(\frac{5}{36}\right)$
$R_H=\frac{36}{5\times 656.1}$
$\frac{1}{{\lambda }_2}=R_H\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$
$\frac{1}{{\lambda }_2}=\frac{36}{5\times 626.1}\left(\frac{3}{16}\right)$
$\lambda =\frac{5\times 656.1\times 16}{3\times 36}=486.0nm$