Question

Does a photon of visible light \((\lambda \approx 400\) to \(700 \mathrm{~nm}\) ) have sufficient energy to excite an electron in a hydrogen atom from the \(n=1\) to the \(n=5\) energy state? from the \(n=2\) to the \(n=6\) energy state?

Short answer

A photon of visible light does not have sufficient energy to excite an electron in a hydrogen atom from the n=1 to the n=5 energy state (\(\Delta E_{1\to 5} = 13.056\,\text{eV}\)) or from the n=2 to the n=6 energy state (\(\Delta E_{2\to 6} = 3.022\,\text{eV}\)). The energy range of visible light is approximately \(1.77\,\text{eV} \leq E_\text{photon} \leq 3.103\,\text{eV}\), which is lower than the required energy for these excitations.

Step-by-step solution

Calculate the initial and final energy levels for both cases

For case 1 (n=1 to n=5): Initial energy level, \(E_1 = -\frac{13.6\,\text{eV}}{(1)^2} = -13.6\,\text{eV}\) Final energy level, \(E_5 = -\frac{13.6\,\text{eV}}{(5)^2} = -0.544\,\text{eV}\) For case 2 (n=2 to n=6): Initial energy level, \(E_2 = -\frac{13.6\,\text{eV}}{(2)^2} = -3.4\,\text{eV}\) Final energy level, \(E_6 = -\frac{13.6\,\text{eV}}{(6)^2} = -0.378\,\text{eV}\)

Calculate the energy differences for both cases

For case 1: Energy difference, \(\Delta E_{1\to 5} = E_5 - E_1 = -0.544\,\text{eV} - (-13.6\,\text{eV}) = 13.056\,\text{eV}\) For case 2: Energy difference, \(\Delta E_{2\to 6} = E_6 - E_2 = -0.378\,\text{eV} - (-3.4\,\text{eV}) = 3.022\,\text{eV}\)

Calculate the energy range of visible light

The energy of a photon can be calculated using the formula: \[E_\text{photon} = \frac{hc}{\lambda}\] Where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the photon. We will calculate the minimum and maximum energy of visible light: Minimum energy (longest wavelength at \(\lambda = 700\,\text{nm}\)): \[E_\text{min} = \frac{h c}{\lambda_\text{max}} = \frac{4.1357 \times 10^{-15}\,\mathrm{eV\,s} \times 2.998\times 10^{8}\,\mathrm{m\,s^{-1}}}{700 \times 10^{-9}\,\mathrm{m}} = 1.77\,\text{eV}\] Maximum energy (shortest wavelength at \(\lambda = 400\,\text{nm}\)): \[E_\text{max} = \frac{h c}{\lambda_\text{min}} = \frac{4.1357 \times 10^{-15}\,\mathrm{eV\,s} \times 2.998\times 10^{8}\,\mathrm{m\,s^{-1}}}{400 \times 10^{-9}\,\mathrm{m}} = 3.103\,\text{eV}\]

Compare the energy differences and photon energy range

Now we'll examine if the energy differences for both cases fall within the energy range of visible light: For case 1: \(\Delta E_{1\to 5} = 13.056\,\text{eV}\) falls outside the visible light energy range (\(1.77\,\text{eV} \leq E_\text{photon} \leq 3.103\,\text{eV}\)), so the photon does not have sufficient energy to excite an electron from n=1 to n=5. For case 2: \(\Delta E_{2\to 6} = 3.022\,\text{eV}\) is slightly above the maximum energy of the visible light (\(3.103\,\text{eV}\)), so the photon does not have sufficient energy to excite an electron from n=2 to n=6. In conclusion, a photon of visible light does not have sufficient energy to excite an electron in a hydrogen atom from the n=1 to the n=5 energy state or from the n=2 to the n=6 energy state.

Key concepts

Energy Levels in the Hydrogen Atom

The hydrogen atom serves as a fundamental example in quantum mechanics due to its simplicity and well-understood spectral characteristics. Regarding energy levels, each electron in a hydrogen atom resides in a specific, quantized state represented by an integer called the principal quantum number, denoted as n. These levels, described by the Bohr model and further refined by quantum mechanics, indicate that the electron has discrete energies rather than a continuous spectrum.

For the hydrogen atom, the energy E_n of an electron in a specific level n can be calculated using the formula: \[E_n = -\frac{13.6\,\text{eV}}{n^2}\].The negative sign indicates that the electron is bound to the atom; the greater the value of n, the less negative the energy becomes, signifying that the electron is less tightly bound and further from the nucleus. An important aspect is that to shift an electron from a lower state (n = 1) to a higher state (n = 5), energy must be absorbed, equal to the difference between the two levels. This particular process is called electron excitation.

Visible Light Photon Energy

Visible light refers to the portion of the electromagnetic spectrum that our human eyes can perceive, with wavelengths roughly between 400 nm (violet) and 700 nm (red). The energy of a visible light photon can be understood through the equation:\[E_\text{photon} = \frac{hc}{\lambda}\].Here, h is the Planck constant, c is the speed of light, and \lambda represents the wavelength of the light. What's interesting is that photons with shorter wavelengths carry more energy than those with longer wavelengths. For instance, a violet photon at around 400 nm is higher in energy than a red photon at about 700 nm. The energy range visible light spans is insufficient to cause transition in a hydrogen atom for electrons jumping from the first to the fifth energy state (n=1 to n=5), or from the second to the sixth state (n=2 to n=6), as we have seen in the exercise.

Quantum Mechanical Model

The quantum mechanical model of the atom is a foundation of modern physics and chemistry, offering a much more accurate description of atoms and their subatomic particles than classical models. Unlike the previous models which depicted electrons in static orbits, the quantum mechanical model interprets the electron as existing in a cloud of probability, called an orbital. Each orbital describes a region in space where there is a high probability of finding an electron.

When energy is absorbed by an electron, it moves to a higher energy orbital, a phenomenon known as excitation. Conversely, when an electron drops to a lower energy level, it emits energy typically in the form of a photon, which might be visible light if the energy change corresponds to the visible spectrum. The quantum mechanical model tells us that not all photon energies can cause the excitation of an electron; only those photons that have energy corresponding to the exact difference between the defined energy levels of the atom can result in an electron transition, as illuminated in the exercise.