Answer:
The wavelength of the visible line in the hydrogen spectrum is 434 nm.
Explanation:
It is given that, the wavelength of the visible line in the hydrogen spectrum that corresponds to n₂ = 5 in the Balmer equation.
For Balmer series, the wave number is given by :
[tex]\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})[/tex]
R is the Rydberg's constant
For Balmer series, n₁ = 2. So,
[tex]\dfrac{1}{\lambda}=1.097\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{5^2})[/tex]
[tex]\lambda=4.34\times 10^{-7}\ m[/tex]
or
[tex]\lambda=434\ nm[/tex]
So, the wavelength of the visible line in the hydrogen spectrum is 434 nm. Hence, this is the required solution.
