Question

The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2. Compute the wavelength of this line for l H and 21. Note: These are the electronic hydrogen and deuterium atoms, not the muonic forms.]

Answer 1

Answer:

$\lambda=550\ nm$

Explanation:

The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2 i.e. $n_i=3$ and $n_f=2$

The wavelength of Hi line of the Balmer series is given by :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_f}-\dfrac{1}{n_i})$

$\dfrac{1}{\lambda}=1.09\times 10^7\times (\dfrac{1}{2}-\dfrac{1}{3})$

$\lambda=5.50\times 10^{-7}\ m$

$\lambda=550\ nm$

So, the wavelength for this line is 550 nm. Hence, this is the required solution.