Answer:
[tex]\lambda=550\ nm[/tex]
Explanation:
The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2 i.e. [tex]n_i=3[/tex] and [tex]n_f=2[/tex]
The wavelength of Hi line of the Balmer series is given by :
[tex]\dfrac{1}{\lambda}=R(\dfrac{1}{n_f}-\dfrac{1}{n_i})[/tex]
[tex]\dfrac{1}{\lambda}=1.09\times 10^7\times (\dfrac{1}{2}-\dfrac{1}{3})[/tex]
[tex]\lambda=5.50\times 10^{-7}\ m[/tex]
[tex]\lambda=550\ nm[/tex]
So, the wavelength for this line is 550 nm. Hence, this is the required solution.
