Question

First ionization energy of hydrogen atom= 2.18 aJ (attojoules). What the frequency and wavelength, in nanometers, of photons capable of just ionizing hydrogen atoms?

Answer 1

The frequency of photons capable of just ionizing hydrogen atoms is approximately 3.29 x 10^15 Hz, and the wavelength is approximately 91 nm.

Step-by-step explanation:

The ionization energy of a hydrogen atom is the minimum energy required to remove an electron from the atom. In this case, the ionization energy is given as 2.18 aJ. To calculate the frequency and wavelength of photons capable of ionizing hydrogen atoms, we can use the formula E = hf, where E is the energy of the photon, h is the Planck's constant (6.626 x 10^-34 Js), and f is the frequency of the photon.

1. First, we can calculate the energy of the photon by converting the ionization energy to joules: 2.18 aJ = 2.18 x 10^-18 J.
2. Next, we can rearrange the formula and solve for f: f = E / h = (2.18 x 10^-18 J) / (6.626 x 10^-34 Js).
3. Plugging in the values, we get: f = 3.29 x 10^15 Hz.
4. To find the wavelength, we can use the equation c = λf, where c is the speed of light (3 x 10^8 m/s) and λ is the wavelength. Rearranging the formula, we get: λ = c / f = (3 x 10^8 m/s) / (3.29 x 10^15 Hz).
5. Plugging in the values, we get: λ ≈ 9.1 x 10^-8 m = 91 nm.

Therefore, the frequency of photons capable of just ionizing hydrogen atoms is approximately 3.29 x 10^15 Hz (or 3.29 THz), and the wavelength is approximately 91 nm.