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 [tex]\mu[/tex]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:
[tex]E=E_{0} Z^2 [\frac{1}{n_1^2}-\frac{1}{n_2^2}][/tex]
, where E is the photon energy, [tex]E_0[/tex] is the energy of the first energy level (-13.6 eV), Z is the atomic number, [tex]n_1[/tex] is the quantum number n of the starting level and [tex]n_2[/tex] the quantum number n of the finishing level. In this case, [tex]n_2=3[/tex], and [tex]n_1=4[/tex], because this excited level is the next in energy to n=3.
Considering that [tex]1 eV= 1.60217662x10^{-19} J[/tex], and using the Planck equation [tex]E=h\nu=\frac{hc}{\lambda}[/tex], you can calculate the wavelenght or the frequency associated to that photon. Values in the order of [tex]\mu[/tex]m in wavelenght belong to the Infrared spectrum, wich can not being seen by humans.
Answer:
1. [tex]E=1.059x10^{-19}J[/tex]
2. [tex]n=4[/tex]
3. The associated wavelength belongs to the infrared spectrum which is invisible for humans.
Explanation:
Hello,
1. At first, according to the equation:
[tex]E=E_oZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})[/tex]
Whereas [tex]E[/tex] is the emitted energy, [tex]E_o[/tex] the first level energy, [tex]Z[/tex] the atomic number, [tex]n_1[/tex] the first level and [tex]n_2[/tex] the second level. In such a way, the closest the [tex]n[/tex]'s are, the least the amount of emitted energy, therefore, [tex]n_1=3[/tex] (based on the statement) and [tex]n_2=4[/tex], thus, the least amount of energy turns out being:
[tex]E=(2.179x10^{-18}J)(1^2)(\frac{1}{3^2}-\frac{1}{4^2})\\E=1.059x10^{-19}J[/tex]
2. Secondly, and based on the first question, the quantum number for the excited state is mandatorily [tex]n=4[/tex]
3. Finally, to substantiate why we cannot observe the emitted photons we apply the following equation:
[tex]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[/tex]
Whereas the obtained wavelength corresponds to the infrared spectrum which is not observable by humans.
Best regards.
