Calculate the wavelength of light emitted when an electron transitions from the n=7 to n=2 energy level in a…

Question

Calculate the wavelength of light emitted when an electron transitions from the n=7 to n=2 energy level in a hydrogen atom using the Rydberg formula.

ScanSolve AI Tutor · Accepted Answer

To find the wavelength of light in the Balmer series, we use the Rydberg formula:

\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

where:

Identify the given values.
For the Balmer series, the electron transitions to the $n_f = 2$ energy level.
The given initial principal quantum number is $n_i = 7$ .
The Rydberg constant $R = 1.097 \times 10^{-2} \text{ nm}^{-1}$ .
Substitute the values into the Rydberg formula.
$\frac{1}{\lambda} = (1.097 \times 10^{-2} \text{ nm}^{-1}) \left( \frac{1}{2^2} - \frac{1}{7^2} \right)$
Calculate the terms inside the parenthesis.
$\frac{1}{\lambda} = (1.097 \times 10^{-2} \text{ nm}^{-1}) \left( \frac{1}{4} - \frac{1}{49} \right)$ $\frac{1}{\lambda} = (1.097 \times 10^{-2} \text{ nm}^{-1}) (0.25 - 0.02040816)$ $\frac{1}{\lambda} = (1.097 \times 10^{-2} \text{ nm}^{-1}) (0.22959184)$
Multiply by the Rydberg constant.
$\frac{1}{\lambda} = 0.0025187 \text{ nm}^{-1}$
Calculate the wavelength $\lambda$ .
$\lambda = \frac{1}{0.0025187 \text{ nm}^{-1}}$ $\lambda \approx 397.02 \text{ nm}$
Comparing this value with the given options, it matches option D.

The final answer is $\text{D. 397.0 nm}$ .

Calculate the wavelength of light emitted when an electron transitions from the n=7 to n=2 energy level in a hydrogen atom using the Rydberg formula.