How much energy is required to ionize a hydrogen atom in its ground (or lowest energy) state?
What wavelength of light contains enough energy in a single photon to ionize a hydrogen atom?
I would do the second part first.
1/lambda = R*(1/N^2 - 1/N^2).
You know the Rydberg constant, the first N is 1 and and N2 is infinity. Solve for lambda.
Then E = hc/lambda
To determine the energy required to ionize a hydrogen atom in its ground state, we need to consider the ionization energy of hydrogen.
1. The ionization energy of hydrogen is approximately 13.6 eV (electron volts).
To calculate the energy required in joules, we can use the conversion factor:
1 eV = 1.602 x 10^-19 J
2. Multiply the ionization energy by the conversion factor:
13.6 eV x 1.602 x 10^-19 J/eV = 2.18 x 10^-18 J
Therefore, the amount of energy required to ionize a hydrogen atom in its ground state is approximately 2.18 x 10^-18 Joules.
Regarding the second question, we can use the equation relating the energy of a single photon to its wavelength:
Energy of a photon (E) = Planck's constant (h) × Speed of light (c) / Wavelength (λ)
To find the minimum wavelength required to ionize a hydrogen atom, we need to equate the energy of the photon to the ionization energy of hydrogen.
1. Substitute the known values into the equation:
2.18 x 10^-18 J = (6.63 x 10^-34 J·s) × (3.00 x 10^8 m/s) / λ
2. Rearrange the equation to solve for the wavelength (λ):
λ = (6.63 x 10^-34 J·s) × (3.00 x 10^8 m/s) / 2.18 x 10^-18 J
3. Calculate the value of λ:
λ ≈ 91.18 nm
Therefore, the wavelength of light containing enough energy in a single photon to ionize a hydrogen atom is approximately 91.18 nanometers (nm).
To determine the energy required to ionize a hydrogen atom in its ground state, we need to consider the ionization energy of hydrogen. The ionization energy is defined as the minimum amount of energy required to completely remove an electron from an atom or ion.
The ionization energy of hydrogen can be calculated using the Rydberg formula:
E = -RH/n^2
Where E is the energy, RH is the Rydberg constant (approximately 2.18 x 10^(-18) Joules), and n is the principal quantum number of the energy level (in this case, n = 1 for the ground state).
Plugging in the values, we can calculate:
E = - (2.18 x 10^(-18) J) / (1^2)
E ≈ -2.18 x 10^(-18) J
The negative sign indicates that energy is released when the electron is removed, as it moves from a lower energy state to a higher energy state. Therefore, the energy required to ionize a hydrogen atom in its ground state is approximately 2.18 x 10^(-18) Joules.
To find the wavelength of light that contains enough energy in a single photon to ionize a hydrogen atom, we can use the equation:
E = hc/λ
Where E is the energy of a photon, h is the Planck's constant (approximately 6.63 x 10^(-34) J·s), c is the speed of light (approximately 3 x 10^8 m/s), and λ is the wavelength.
Plugging in the values, we can rearrange the equation to solve for λ:
λ = hc/E
Substituting the value of E for ionization of hydrogen (approximately 2.18 x 10^(-18) J), we get:
λ = (6.63 x 10^(-34) J·s)(3 x 10^8 m/s) / (2.18 x 10^(-18) J)
λ ≈ 9.09 x 10^(-8) meters
Therefore, the wavelength of light that contains enough energy in a single photon to ionize a hydrogen atom is approximately 9.09 x 10^(-8) meters, which corresponds to ultraviolet (UV) light.