Question

the energy ( e ) of the electron in a hydrogen atom can be calculated from the bohr formula:

( e = -\frac{r_y}{n^2} )

in this equation ( r_y ) stands for the rydberg energy, and ( n ) stands for the principal quantum number of the orbital that holds the electron. (you can find the value of the rydberg energy using the data button on the aleks toolbar.)

calculate the wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron from an orbital with ( n = 9 ) to an orbital with ( n = 4 ). round your answer to 3 significant digits.

Explanation:

Step1: Calculate the energy of the initial and final states

The Rydberg energy \(R_y = 2.18\times10^{-18}\space J\)
For \(n = 9\), \(E_9=-\frac{R_y}{n^{2}}=-\frac{2.18\times 10^{-18}\space J}{9^{2}}=-\frac{2.18\times 10^{-18}\space J}{81}\approx - 2.69\times10^{-20}\space J\)
For \(n = 4\), \(E_4=-\frac{R_y}{n^{2}}=-\frac{2.18\times 10^{-18}\space J}{4^{2}}=-\frac{2.18\times 10^{-18}\space J}{16}=- 1.3625\times10^{-19}\space J\)

Step2: Calculate the energy change \(\Delta E\)

\(\Delta E=E_4 - E_9\)
\(\Delta E=-1.3625\times 10^{-19}\space J-(-2.69\times 10^{-20}\space J)\)
\(\Delta E=-1.3625\times 10^{-19}+2.69\times 10^{-20}\space J\)
\(\Delta E=-(1.3625 - 0.269)\times10^{-19}\space J=- 1.0935\times10^{-19}\space J\) (The negative sign indicates that energy is emitted)
\(\vert\Delta E\vert = 1.0935\times 10^{-19}\space J\)

Step3: Use the formula \(\Delta E=\frac{hc}{\lambda}\) to find \(\lambda\)

We know that \(h = 6.626\times10^{-34}\space J\cdot s\) and \(c = 3\times10^{8}\space m/s\)
From \(\lambda=\frac{hc}{\Delta E}\)
\(\lambda=\frac{6.626\times 10^{-34}\space J\cdot s\times3\times 10^{8}\space m/s}{1.0935\times 10^{-19}\space J}\)
\(\lambda=\frac{19.878\times 10^{-26}}{1.0935\times 10^{-19}}\space m\)
\(\lambda\approx1.82\times10^{-6}\space m\)
Since \(1\space m = 10^{6}\space\mu m\), \(\lambda = 1.82\space\mu m\)

Answer:

\(1.82\space\mu m\)