Question

What is the smallest energy photon that can be absorbed by a hydrogen atom in its ground state?

Short answer

The smallest energy photon that can be absorbed by a hydrogen atom in its ground state has an energy of 10.2 eV, which corresponds to a wavelength of 121.5 nm in the ultraviolet range of the electromagnetic spectrum.

Step-by-step solution

Understanding hydrogen atom energy levels

To determine the smallest energy photon absorbed by a hydrogen atom, we need to understand its energy levels. The energy levels of a hydrogen atom can be described by the following equation: E(n) = -\frac{13.6 \,\text{eV}}{n^2} Where E(n) is the energy of the electron when it's in the nth energy level, and n is the principal quantum number, an integer ranging from 1 (ground state) to infinity.

Determining the transition

In order to determine the smallest energy photon absorbed, we need to look at the smallest transition from the ground state (n = 1) to the next energy level (n = 2). Calculating the energy difference between these two levels will yield the energy of the smallest photon, which can be absorbed.

Calculating energy difference between levels

We need to calculate the energy difference between the ground state (n = 1) and excited state (n = 2). We can do this using the energy equation we obtained in Step 1. Calculate the energy levels for both n = 1 and n = 2: E(1) = -\frac{13.6 \,\text{eV}}{1^2} = -13.6 \,\text{eV} E(2) = -\frac{13.6 \,\text{eV}}{2^2} = -3.4 \,\text{eV} Now find the energy difference (ΔE) between these levels: ΔE = E(2) - E(1) = -3.4 \,\text{eV} - (-13.6 \,\text{eV}) = 10.2 \,\text{eV}

Converting energy difference to photon energy

The energy difference calculated in the previous step corresponds to the energy of the smallest photon that can be absorbed by the hydrogen atom in its ground state. The energy of a photon is given by: E_{\text{photon}} = hf Where E_{\text{photon}} is the photon energy, h is Planck's constant (approximately 4.135667696 x 10^{-15} eV·s), and f is the frequency of the photon. We can rewrite the equation in terms of wavelength (λ) using the speed of light (c): E_{\text{photon}} = \frac{hc}{\lambda} Now we can solve for the smallest wavelength (λ), which corresponds to the smallest energy photon absorbed: \lambda = \frac{hc}{E_{\text{photon}}} = \frac{(4.135667696 \times 10^{-15} \,\text{eV·s}) (3 \times 10^8\, \text{m/s})}{10.2 \,\text{eV}} = 1.215 \times 10^{-7} \,\text{m} This is equivalent to 121.5 nm, which is in the ultraviolet range of the electromagnetic spectrum.

Conclusion

The smallest energy photon that can be absorbed by a hydrogen atom in its ground state has an energy of 10.2 eV and corresponds to a wavelength of 121.5 nm.