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Q. What is the maximum wavelength of line of Balmer series of hydrogen spectrum?
(R = 1.09 × 107 m-1)

AIIMSAIIMS 2018Structure of Atom

Solution:

For maximum wavelength in the Balmer series, n2 = 3 and n1 = 2.

\Rightarrow \quad \frac{1}{\lambda}=1.09\times10^{7}\times1\left(\frac{1}{4}-\frac{1}{9}\right)
\quad\quad\quad\ =1.09\times10^{7}\times1\left(\frac{5}{36}\right)
\Rightarrow \quad \lambda=1.09\times10^{7}\times1\left(\frac{5}{36}\right)
\quad \quad \quad = 6.60 × 10^{-7} m = 660 nm