Explanation:
The given data is as follows.
      [tex]n_{1}[/tex] = 7,      [tex]n_{2}[/tex] = 3
       Z for H = 1
According to Reydberg's equation, we will calculate the energy emitted by the photon as follows.
     [tex]\Delta E = -2.178 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}][/tex]
         = [tex]-2.178 \times 10^{-18} J \times (1)^{2}[\frac{1}{(3)^{2}} - \frac{1}{(7)^{2}}][/tex]
         = [tex]-2.178 \times 10^{-18} J \times (0.11 - 0.02)[/tex]
         = [tex]-1.96 \times 10^{-19} J[/tex]  Â
The negative sign indicates that energy is released in the process.
Thus, we can conclude that energy (in J) of the emitted photon is [tex]-1.96 \times 10^{-19} J[/tex]. Â
