Question

Consider the emission spectrum of singly ionized helium (He^). Find the longest three wavelengths for the series in which the electron makes a transition from a higher excited state to the first excited state (not the ground state).

Short answer

The longest three wavelengths are: - λ₁ = 656.1 nm - λ₂ = 981.0 nm - λ₃ = 1,156.0 nm

Step-by-step solution

Understand the Rydberg formula for hydrogen-like ions

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 \(n_1 = 2\).

Calculate the longest three wavelengths

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 \(n_1 = 2\): $$\frac{1}{\lambda_3} = R_H \cdot 2^2 \cdot \left(\frac{1}{2^2} - \frac{1}{5^2}\right)$$

Calculate the wavelengths

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 \(\lambda_3\): $$\lambda_3 = \frac{1}{R_H \cdot 2^2 \cdot \left(\frac{1}{2^2} - \frac{1}{5^2}\right)}$$

Finalize the solution

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: $$\lambda_1 = 656.1 \, \text{nm}$$ $$\lambda_2 = 981.0 \, \text{nm}$$ $$\lambda_3 = 1,156.0 \, \text{nm}$$