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Calculate wavelength of light emitted when an electron in a hydrogen atom undergoes sa transition from n=4 to n=2

User GBouffard
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Final answer:

Using the Rydberg formula, the wavelength of light emitted from the transition of an electron from n=4 to n=2 in a hydrogen atom is calculated to be approximately 486 nm, which is in the visible range of light.

Step-by-step explanation:

To calculate the wavelength of light emitted when an electron in a hydrogen atom undergoes a transition from n=4 to n=2, we can use 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 of the emitted light
  • \(R_H\) is the Rydberg constant (\(1.097 \times 10^7 m^{-1}\))
  • \(n_1\) and \(n_2\) are the principal quantum numbers of the initial and final energy levels, respectively

Plugging the values into the formula:

\[\frac{1}{\lambda} = (1.097 \times 10^7 m^{-1}) \left(\frac{1}{2^2} - \frac{1}{4^2}\right)\]

\[\frac{1}{\lambda} = (1.097 \times 10^7 m^{-1}) \left(\frac{1}{4} - \frac{1}{16}\right)\]

\[\frac{1}{\lambda} = (1.097 \times 10^7 m^{-1}) \left(\frac{4}{16} - \frac{1}{16}\right)\]

\[\frac{1}{\lambda} = (1.097 \times 10^7 m^{-1}) \left(\frac{3}{16}\right)\]

\[\frac{1}{\lambda} = 2.055 \times 10^6 m^{-1}\]

\[\lambda = \frac{1}{2.055 \times 10^6 m^{-1}}\]

\[\lambda \approx 4.86 \times 10^{-7} m\]

The wavelength of light emitted by this transition is approximately 486 nm, which falls within the visible light spectrum, specifically within the blue range.

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