Energy of the electron in the $n^{\text{th}}$ state of an atom Here, $Z$ is the atomic number of the atom.For hydrogen atom, $Z=1$Energy required to excite an atom from initial state $(n_i)$ to final state $(n_f)$,$E = -13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{eV}$$\Rightarrow \frac{-13.6}{n_f^2} + \frac{13.6}{n_i^2} = 12.9$ This energy must be equal to or less than the energy of the incident electron beam.$\Rightarrow 13.6 - 12.9 = \frac{13.6}{n_f^2} \; [\because n' = 1]$$\Rightarrow n_f = 4.4$State cannot be a fraction number.$\Rightarrow n_f = 4$Hence, the hydrogen atom would be excited up to $4^{\text{th}}$ energy level.Rydberg's formula for the spectrum of the hydrogen atom is given by:$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$Here, $\lambda$ is the wavelengthRydberg's canstant, $R = 1.097 \times 10^7 \text{m}^{-1}$For the first member of the Paschen series$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{4^2} \right)$$\lambda = 18752.4 \text{Å}$For the first member of Balmer series $n_1 = 2, n_2 = 3$$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$$\lambda = 6563.3 \text{Å}$