Questions: Calculate the wavelength of light (in nm) for an electron in a H atom going between the n=2 and n=3 energy levels. Report your answer to the nearest nm. RH=1.097 x 10^7 m^-1 (Since we are dealing with wavelengths, these must always be positive numbers!)

Solution Steps

We need to calculate the wavelength of light emitted or absorbed when an electron transitions between the \( n=2 \) and \( n=3 \) energy levels in a hydrogen atom. This involves using the Rydberg formula for hydrogen:

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

where \( \lambda \) is the wavelength, \( R_H = 1.097 \times 10^7 \, \text{m}^{-1} \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

For the transition from \( n=2 \) to \( n=3 \):

• \( n_1 = 2 \)
• \( n_2 = 3 \)

Substitute the values into the Rydberg formula:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Calculate the terms inside the parentheses:

\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \] \[ \frac{1}{3^2} = \frac{1}{9} \approx 0.1111 \]

Subtract these values:

\[ 0.25 - 0.1111 = 0.1389 \]

Now, substitute back into the formula:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.1389 \]

Calculate the right-hand side:

\[ \frac{1}{\lambda} \approx 1.524 \times 10^6 \, \text{m}^{-1} \]

To find \( \lambda \), take the reciprocal:

\[ \lambda = \frac{1}{1.524 \times 10^6} \approx 6.562 \times 10^{-7} \, \text{m} \]

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

\[ \lambda \approx 6.562 \times 10^{-7} \times 10^9 = 656.2 \, \text{nm} \]

Round to the nearest nanometer:

\[ \lambda \approx 656 \, \text{nm} \]

Final Answer