Question

Does a photon of visible light \((\lambda=400 \text { to } 700 \mathrm{nm}\) ) 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, with wavelengths in the range of \(400\,\text{nm}\) to \(700\,\text{nm}\), has an energy range of approximately \(1.771\,\text{eV}\) to \(3.103\,\text{eV}\). To excite an electron in a hydrogen atom from the \(n=1\) to the \(n=5\) energy state, \(13.056\,\text{eV}\) are required, while \(3.023\,\text{eV}\) are needed to excite it from the \(n=2\) to the \(n=6\) energy state. In both cases, the energy of a visible light photon is insufficient to excite the electron to these higher energy states.

Step-by-step solution

Calculate the energy needed to excite an electron from n=1 to n=5

In a hydrogen atom, the energy of the electron in various energy levels (n) is given by the formula: \[ E_n = -\dfrac{13.6\,\text{eV}}{n^2} \] We need to calculate the energy difference between n=5 and n=1, which gives the required energy to excite the electron from the ground state to the n=5 state: \[ \Delta E_{1 \to 5} = E_5 - E_1 \] Using the energy formula, we can calculate the energy of the 1st and 5th energy levels: \[ E_1 = -\dfrac{13.6\,\text{eV}}{1^2} = -13.6\,\text{eV} \] \[ E_5 = -\dfrac{13.6\,\text{eV}}{5^2} = -\dfrac{13.6\,\text{eV}}{25} = -0.544\,\text{eV} \] Now we can find the energy difference: \[ \Delta E_{1 \to 5} = E_5 - E_1 = -0.544\,\text{eV} - (-13.6\,\text{eV}) = 13.056\,\text{eV} \]

Calculate the energy needed to excite an electron from n=2 to n=6

Similarly, we need to calculate the energy difference between n=6 and n=2: \[ \Delta E_{2 \to 6} = E_6 - E_2 \] \[ E_2 = -\dfrac{13.6\,\text{eV}}{2^2} = -\dfrac{13.6\,\text{eV}}{4} = -3.4\,\text{eV} \] \[ E_6 = -\dfrac{13.6\,\text{eV}}{6^2} = -\dfrac{13.6\,\text{eV}}{36} = -0.377\,\text{eV} \] Now we can find the energy difference: \[ \Delta E_{2 \to 6} = E_6 - E_2 = -0.377\,\text{eV} - (-3.4\,\text{eV}) = 3.023\,\text{eV} \]

Calculate the energy range of visible light photons

Visible light has wavelengths in the range of 400 nm to 700 nm. We need to find the energy of these photons. The energy of a single photon is given by the formula: \[ E = h \nu = \dfrac{hc}{\lambda} \] where h is the Planck's constant ( \(h = 6.63 \times 10^{-34}\,\text{Js}\)), c is the speed of light ( \(c = 3.00 \times 10^8\,\text{m/s}\)), and λ is the wavelength of the photon. We can calculate the energy range for visible light photons: \[ E_{400} = \dfrac{6.63 \times 10^{-34}\,\text{Js} \times 3.00 \times 10^8\,\text{m/s}}{400 \times 10^{-9}\,\text{m}} = 4.964 \times 10^{-19}\,\text{J} \] \[ E_{700} = \dfrac{6.63 \times 10^{-34}\,\text{Js} \times 3.00 \times 10^8\,\text{m/s}}{700 \times 10^{-9}\,\text{m}} = 2.834 \times 10^{-19}\,\text{J} \] To make the comparison easier, we can convert these energies to eV: \[ E_{400\,\text{eV}} = \dfrac{4.964 \times 10^{-19}\,\text{J}}{1.6 \times 10^{-19}\,\text{J/eV}} = 3.103\,\text{eV} \] \[ E_{700\,\text{eV}} = \dfrac{2.834 \times 10^{-19}\,\text{J}}{1.6 \times 10^{-19}\,\text{J/eV}} = 1.771\,\text{eV} \] The energy range for visible light photons is 1.771 eV to 3.103 eV.

Compare the energy ranges

Now, we will compare the required energies to excite the electron from n=1 to n=5 and from n=2 to n=6 with the energy range of visible light photons. For n=1 to n=5, the required energy is 13.056 eV, which is higher than the energy range of visible light photons (1.771 eV to 3.103 eV). Hence, a photon of visible light does not have sufficient energy to excite an electron from n=1 to n=5. For n=2 to n=6, the required energy is 3.023 eV, which is also higher than the energy range of visible light photons. Therefore, a photon of visible light does not have sufficient energy to excite an electron from n=2 to n=6 either.

Key concepts

Photon energy

The concept of photon energy is key to understanding how electrons in atoms are excited from one energy level to another. Photons are particles of light, and each has an energy determined by its frequency \[ E = h u \]or its wavelength \[ E = \dfrac{hc}{\lambda} \]where:

• \( h = 6.63 \times 10^{-34}\,\text{Js} \) is Planck's constant,
• \( c = 3.00 \times 10^8\,\text{m/s} \) is the speed of light,
• \( \lambda \) is the wavelength of the photon.

This equation tells us that energy is inversely proportional to wavelength: as the wavelength decreases, the photon energy increases. By calculating the energy of photons within the visible light spectrum (400 nm to 700 nm), we can see whether these photons have enough energy to move electrons between energy levels in a hydrogen atom.

Visible light wavelength

Visible light occupies a small portion of the electromagnetic spectrum, defined by wavelengths from 400 nm to 700 nm. This range is what the human eye can perceive, and it includes all the colors from violet to red. The energy that corresponds to this range can be calculated using the relationship between energy and wavelength. For 400 nm light, a photon has energy approximately 3.103 eV, and for 700 nm light, it's about 1.771 eV. This energy span shows that visible light photons vary significantly in energy across their range. Understanding this helps us determine whether these photons can excite electrons to higher levels in the atom, which is a crucial aspect of phenomena like absorption and emission of light.

Energy transitions

Energy transitions in atoms occur when an electron jumps from one energy level to another. In a hydrogen atom, these levels are designated by the principal quantum number, \( n \), and are quantized: \[ E_n = -\dfrac{13.6\,\text{eV}}{n^2} \]To move an electron between these levels, energy matching the difference between these levels must be absorbed or emitted.

• Transition from \(n=1\) to \(n=5\) requires 13.056 eV
• Transition from \(n=2\) to \(n=6\) requires 3.023 eV

Comparing these required energies to those available from visible light, we find that visible light photons (1.771 eV to 3.103 eV) often do not provide sufficient energy for such transitions. Therefore, other sources of radiation with shorter wavelengths and higher energies, like ultraviolet light, are more effective in exciting electrons in a hydrogen atom.