Suppose that at least 13.0 eV is needed to free an electron from a particular element, i.e., to ionize the atom. What is the lowest frequency photon which can accomplish this? And, what is the corresponding wavelength for that photon? Give frequency for answer 1 and wavelength for answer 2.
Suppose that at least 13.0 eV is needed to free an electron from a particular element, i.e., to ionize the atom. What is the lowest frequency photon which can accomplish this?
To find the lowest frequency photon that can free an electron from an atom, we need to convert the energy of 13.0 eV to joules.
1 eV is equal to 1.602 x 10^-19 joules, so 13.0 eV is equal to 13.0 x 1.602 x 10^-19 joules.
Once we have the energy in joules, we can use the equation E = hf, where E is the energy, h is Planck's constant (6.626 x 10^-34 J s), and f is the frequency of the photon.
First, we rearrange the equation to solve for frequency: f = E/h.
Plugging in the values:
f = (13.0 x 1.602 x 10^-19 joules) / (6.626 x 10^-34 J s).
Calculating the result:
f ≈ 3.096 x 10^15 Hz.
So, the lowest frequency photon that can free an electron from the atom is approximately 3.096 x 10^15 Hz.
To find the corresponding wavelength, we can use the formula c = λf, where c is the speed of light (approximately 3.0 x 10^8 m/s), λ is the wavelength, and f is the frequency.
First, rearrange the equation to solve for wavelength: λ = c/f.
Plugging in the values:
λ = (3.0 x 10^8 m/s) / (3.096 x 10^15 Hz).
Calculating the result:
λ ≈ 9.69 x 10^-8 meters.
So, the corresponding wavelength for the lowest frequency photon is approximately 9.69 x 10^-8 meters.