Question

What will be the longest wavelength line in the Balmer series of spectrum of H atom?

• A. $546nm$
• B. $656nm$
• C. $566nm$
• D. $556nm$

Answer 1

The correct option is B $656nm$

For single electron species,

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

where, R = Rydberg constant = $1.097\times {10}^7\times {10}^{-9}n{m}^{-1}=1.097\times {10}^{-2}n{m}^{-1}$

For Hydrogen atom, Z = Atomic number = 1

For Balmer series, $n_1$ = 2 and for longest wavelength in Balmer series, minimum energy transition is to be considered because wavelength is inversely proportional to Energy.

So, $n_2$ = 3

$\therefore \frac{1}{\lambda }=1.097\times {10}^{-2}n{m}^{-1}\times 1^2(\frac{1}{2^2}-\frac{1}{3^2})$

$\Rightarrow \lambda =656nm$