Example 8 Atomic hydrogen emitted only three spectral lines during the transition from the excited to the ground state Find a wavelength of the maximum frequency among them (nm)

To find the wavelength of the maximum frequency among the spectral lines emitted by atomic hydrogen during the transition from the excited state to the ground state, we can use the Rydberg formula:<br /><br />1/λ = R_H * (1/n1^2 - 1/n2^2)<br /><br />where λ is the wavelength, R_H is the Rydberg constant, n1 is the principal quantum number of the lower energy level (ground state), and n2 is the principal quantum number of the higher energy level (excited state).<br /><br />For the maximum frequency, we need to find the transition with the smallest wavelength. The smallest wavelength corresponds to the largest energy difference between the two energy levels. In this case, the transition from the n=2 energy level to the n=1 energy level (the ground state) will have the largest energy difference.<br /><br />Using the Rydberg formula, we can calculate the wavelength of the maximum frequency:<br /><br />1/λ = R_H * (1/1^2 - 1/2^2)<br />1/λ = R_H * (1 - 1/4)<br />1/λ = R_H * (3/4)<br /><br />Rearranging the equation, we get:<br /><br />λ = 4 / (3 * R_H)<br /><br />Substituting the value of the Rydberg constant (R_H = 1.097373 x 10^7 m^-1), we can calculate the wavelength:<br /><br />λ = 4 / (3 * 1.097373 x 10^7 m^-1)<br />λ ≈ 1.021 x 10^-7 m<br /><br />Converting this to nanometers (1 m = 10^9 nm):<br /><br />λ ≈ 102.1 nm<br /><br />Therefore, the wavelength of the maximum frequency among the spectral lines emitted by atomic hydrogen during the transition from the excited state to the ground state is approximately 102.1 nm.