91影视

First, we will calculate the energy of the emission spectral line using the given wavelength. The energy of a photon is given by: $$E = h \cdot c / \lambda$$ Where: - \(\lambda\) is the wavelength of the emission spectral line - \(E\) is the energy - \(h\) is the Planck's constant (\(6.63 \times10^{-34} \ \mathrm{J \cdot s}\) ) - \(c\) is the speed of light (\(3.00 \times10^{8} \ \mathrm{m/s}\)) First, convert the given wavelength from nanometers to meters: $$\lambda = 253.4 \times10^{-9} \ \mathrm{m}$$ Then, calculate the energy: $$E = \frac{6.63 \times10^{-34} \ \mathrm{J \cdot s} \cdot 3.00 \times10^{8} \ \mathrm{m/s}}{253.4 \times10^{-9} \ \mathrm{m}}$$ $$E = 7.87 \times10^{-19} \ \mathrm{J}$$

Now, we will use the Rydberg formula to solve for the principal quantum number of the lower-energy state. Rearranging the equation so that \(n^2\) is the subject, we have: $$n^2 = \frac{hcRZ^2}{E}$$ Where: - \(E\) is the change in energy from Step 1 (\(7.87 \times10^{-19}\ \mathrm{J}\)) - \(Z\) is the nuclear charge (4 for \(\mathrm{Be}^{3+}\)) - \(R\) is the Rydberg constant for hydrogen (\(1.097 \times10^{7} \ \mathrm{m^{-1}}\)) We need to find \(n^2\) and then take the square root of the answer to find the final principal quantum number, \(n\). We have: $$n^2 = \frac{(6.63 \times10^{-34}\ \mathrm{J \cdot s}) \cdot(3.00 \times10^{8} \ \mathrm{m/s}) \cdot (1.097 \times10^{7} \ \mathrm{m^{-1}}) \cdot (4)^2}{7.87 \times10^{-19} \ \mathrm{J}}$$ $$n^2 \approx 20.52$$ Now, take the square root to determine the principal quantum number, rounding to the nearest whole number: $$n = \sqrt{20.52} \approx 4.5$$ Since the quantum number must be a whole number, we round it up to the nearest integer, \(n = 5\). However, this would mean that the initial and final quantum number would be the same, which is not possible for an electronic transition. Therefore, we round it down to the nearest integer, \(n=4\). The principal quantum number of the lower-energy state corresponding to this emission is 4.