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