Use Balmer’s formula to calculate (a) the wavelength, (b) the frequency, and (c) the photon energy for the H_ɣ line of the Balmer series for hydrogen.

According to de Broglie, Light has dual nature as in some situations it behaves like waves and in others like particles. Electrons, which sometimes exhibits particle nature may also behave like wave in some situations.

When a particle acts like a wave, it carries a wavelength which is known as de Broglie wavelength.

Wavelength of the emitted photon for Balmer series is given by

$$\begin{gathered} \frac{1}{\lambda }=R\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \\ Rydbergcons\tan t,R=1.097*{10}^7{m}^{-1} \\ BalmerSeries\left(n_2=2\right) \\ \frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{n_1^2}\right) \end{gathered}$$

The wavelength for the Balmer series is given by

$$\begin{gathered} \frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{n_1^2}\right) \\ =\left(1.097*{10}^7{m}^{-1}\right)\left(\frac{1}{2^2}-\frac{1}{5^2}\right) \\ =434nm \end{gathered}$$

Therefore, the wavelength for Balmer series is 434 nm.

The frequency for the Balmer series is given by

$f=\frac{c}{\lambda }=\frac{3*{10}^{8}m/s}{4.34*{10}^{-7}m}=6.91*{10}^{14}Hz$

Therefore, the frequency for Balmer series is 6.91 x 10¹⁴ Hz.

The energy of photon for the Balmer series is given by

$$\begin{gathered} E=hf \\ =\left(4.136*{10}^{-15}eVs\right)\left(6.91*{10}^{14}{s}^{-1}\right) \\ =2.86eV \end{gathered}$$

Therefore, the energy of photon for Balmer series is 2.86 eV.