91影视

The general Rydberg formula for hydrogen-like ions is given by: $$\frac{1}{\lambda} = R_H \cdot Z^2 \cdot \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ where \(\lambda\) is the wavelength, \(R_H\) is the Rydberg constant for hydrogen (\(R_H = 1.097 \times 10^7 \, \text{m}^{-1}\)), \(Z\) is the atomic number, \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level. In our case, we are dealing with singly ionized helium (He^), so \(Z = 2\). We are also considering transitions to the first excited state, so \(n_1 = 2\).

We will now find the longest three wavelengths by using the Rydberg formula. To find the longest wavelengths, we need to consider the transitions with the lowest values of \(n_2\). The longest wavelength corresponds to the transition from \(n_2 = 3\) to \(n_1 = 2\): $$\frac{1}{\lambda_1} = R_H \cdot 2^2 \cdot \left(\frac{1}{2^2} - \frac{1}{3^2}\right)$$ The second longest wavelength corresponds to the transition from \(n_2 = 4\) to \(n_1 = 2\): $$\frac{1}{\lambda_2} = R_H \cdot 2^2 \cdot \left(\frac{1}{2^2} - \frac{1}{4^2}\right)$$ The third longest wavelength corresponds to the transition from \(n_2 = 5\) to \(n_1 = 2\): $$\frac{1}{\lambda_3} = R_H \cdot 2^2 \cdot \left(\frac{1}{2^2} - \frac{1}{5^2}\right)$$

Now, we will solve for \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\): For \(\lambda_1\): $$\lambda_1 = \frac{1}{R_H \cdot 2^2 \cdot \left(\frac{1}{2^2} - \frac{1}{3^2}\right)}$$ For \(\lambda_2\): $$\lambda_2 = \frac{1}{R_H \cdot 2^2 \cdot \left(\frac{1}{2^2} - \frac{1}{4^2}\right)}$$ For \(\lambda_3\): $$\lambda_3 = \frac{1}{R_H \cdot 2^2 \cdot \left(\frac{1}{2^2} - \frac{1}{5^2}\right)}$$

By plugging in the values of \(R_H = 1.097 \times 10^7 \, \text{m}^{-1}\) and solving for \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\). We get: $$\lambda_1 = 6.561 \times 10^{-7} \, \text{m}$$ $$\lambda_2 = 9.810 \times 10^{-7} \, \text{m}$$ $$\lambda_3 = 1.156 \times 10^{-6} \, \text{m}$$ So, the longest three wavelengths for the emission spectrum of singly ionized helium (He^) when the electron makes a transition from higher excited states to the first excited state are: $$\lambda_1 = 656.1 \, \text{nm}$$ $$\lambda_2 = 981.0 \, \text{nm}$$ $$\lambda_3 = 1,156.0 \, \text{nm}$$