Question

Use the Rydberg equation to calculate the wavelength (in nm) of the photon emitted when a hydrogen atom undergoes a transition from n = 5 to n = 2.

Answer 1

Answer:

The answer is 434nm

Explanation:

The Rydberg equation is an empirical relationship expressed by Balmer and Rydberg which is stated as:

1/λ =      $R_{H} (\frac{1}{n_{f} ^{2} }-\frac{1}{n_{i} ^{2} } )$.............................. (1)

where λ is the wavelength, $R_{H}$ is the Rydberg constant equal to $1.097 x 10^{7}m^{-1}$, n is the transition level number, the subscript f and i are the final and initial levels respectively. Therefore for final transition n = 2 and for initial transition n = 5.

Making substitutions into equation (1), gives

1/λ =  $1.097 x 10^{7} (\frac{1}{2^{2} }-\frac{1}{5^{2} })$

   = $1.097 x 10^{7} (\frac{1}{4} - \frac{1}{25} )$

     = $1.097 x 10^{7} (0.25 - 0.04)$

   = $1.097 x 10^{7} x 0.21$

   = 2303700

∴ λ = $\frac{1}{2303700}$ $= 4.34 x 10^{-7}m$

Converting to nm, we have

  λ   = $\frac{4.34 x 10^{-7} }{10^{-9} } = 434nm$

Therefore, the wavelength of the emitted photon is 434nm

Answer 2

By using the Rydberg equation, the wavelength (in nm) of the photon emitted is 434 nanometer.

Given the following data:

Initial transition = 5

Final transition = 2

Rydberg constant = $1.090 \times 10^7 \;m^{-1}$

To calculate the wavelength (in nm) of the photon emitted by using the Rydberg equation:

Mathematically, the Rydberg equation is given by the formula:

$1/\lambda = R(\frac{1}{n_f^2} - \frac{1}{n_f^2} )$

Where:

• $\lambda$ is the wavelength.

• R is the Rydberg constant.

• $n_f$ is the final transition.

• $n_i$ is the initial transition.

Substituting the given parameters into the formula, we have;

$1/\lambda = 1.090 \times 10^7 (\frac{1}{2^2} - \frac{1}{5^2} )\\\\1/\lambda = 1.090 \times 10^7 (\frac{1}{4} - \frac{1}{25} )\\\\1/\lambda = 1.090 \times 10^7 (0.25 - 0.04)\\\\1/\lambda = 1.090 \times 10^7 \times 0.21\\\\1/\lambda = 2,289,000\\\\\lambda = \frac{1}{2,289,000} \\\\\lambda = 4.37 \times 10^{-7} \; meters$

To nanometer:

$\lambda = \frac{4.37 \times 10^{-7}}{10^{-9}}\\\\\lambda = 434 \;nanometer$

Wavelength = 434 nanometer.

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