Question

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

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

Answer 1

The correct option is C $486.0nm$

For first line of Balmer series,

$1/\lambda =R(1/2^2-1/3^2)=1/656.1$

For second line of series,

$1/{\lambda }_2=R(1/2^2-1/4^2)$

By putting value of R,

${\lambda }_2=486.0$ nm