Question

What is the least amount of energy that can be emitted by an excited electron in a hydrogen atom falling from an excited state directly to the n = 3 state? What is the quantum number n for the excited state? Humans cannot visually observe the photons emitted in this process. Why not?

Answer 1

Answer:

1. $E=1.059x10^{-19}J$

2. $n=4$

3. The associated wavelength belongs to the infrared spectrum which is invisible for humans.

Explanation:

Hello,

1. At first, according to the equation:

$E=E_oZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Whereas $E$ is the emitted energy, $E_o$ the first level energy, $Z$ the atomic number, $n_1$ the first level and $n_2$ the second level. In such a way, the closest the $n$'s are, the least the amount of emitted energy, therefore, $n_1=3$ (based on the statement) and $n_2=4$, thus, the least amount of energy turns out being:

$E=(2.179x10^{-18}J)(1^2)(\frac{1}{3^2}-\frac{1}{4^2})\\E=1.059x10^{-19}J$

2. Secondly, and based on the first question, the quantum number for the excited state is mandatorily $n=4$

3. Finally, to substantiate why we cannot observe the emitted photons we apply the following equation:

$E=\frac{hc}{\lambda} \\\lambda=\frac{hc}{E}=\frac{(6.62607004x10^{-34} m^2 kg / s)(299 792 458m/s)}{1.059x10^{-19}m^2 kg / s^2} \\\lambda=1.88x10^{-6}m$

Whereas the obtained wavelength corresponds to the infrared spectrum which is not observable by humans.

Best regards.

Answer 2

Answer:

The least amount of energy emitted in this case is 0.6 eV.

The corresponding quantum number n would be n=4.

The wavelenght asociated to the emitted photon would be 2.06 $\mu$m, corresponding to the Infrared spectrum.

Explanation:

For calculating the energy of an electron emitted/absorbed in an electronic transition of the hydrogen atom, the next equation from the Bohr model can be used:

$E=E_{0} Z^2 [\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

, where E is the photon energy, $E_0$ is the energy of the first energy level (-13.6 eV), Z is the atomic number, $n_1$ is the quantum number n of the starting level and $n_2$ the quantum number n of the finishing level. In this case, $n_2=3$, and $n_1=4$, because this excited level is the next in energy to n=3.

Considering that $1 eV= 1.60217662x10^{-19} J$, and using the Planck equation $E=h\nu=\frac{hc}{\lambda}$, you can calculate the wavelenght or the frequency associated to that photon. Values in the order of $\mu$m in wavelenght belong to the Infrared spectrum, wich can not being seen by humans.