Questions: Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transition from the energy level n=7 to the level n=1. λ= nm

Solution Steps

The wavelength of light emitted when an electron transitions between energy levels in a hydrogen atom can be calculated using the Rydberg formula:

\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:

• \(\lambda\) is the wavelength of the emitted light,
• \(R_H\) is the Rydberg constant for hydrogen, approximately \(1.0973 \times 10^7 \, \text{m}^{-1}\),
• \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and higher energy levels, respectively.

For the transition from \(n=7\) to \(n=1\), we have:

• \(n_1 = 1\)
• \(n_2 = 7\)

Substitute the values into the Rydberg formula:

\[ \frac{1}{\lambda} = 1.0973 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{7^2} \right) \]

Calculate the expression inside the parentheses:

\[ \frac{1}{1^2} - \frac{1}{7^2} = 1 - \frac{1}{49} = \frac{49}{49} - \frac{1}{49} = \frac{48}{49} \]

Now, substitute back into the formula:

\[ \frac{1}{\lambda} = 1.0973 \times 10^7 \times \frac{48}{49} \]

Calculate \(\frac{1}{\lambda}\):

\[ \frac{1}{\lambda} = 1.0758 \times 10^7 \, \text{m}^{-1} \]

Finally, calculate \(\lambda\):

\[ \lambda = \frac{1}{1.0758 \times 10^7} \approx 9.292 \times 10^{-8} \, \text{m} \]

Convert meters to nanometers (1 m = \(10^9\) nm):

\[ \lambda \approx 92.92 \, \text{nm} \]

Final Answer