Suppose that at least 14.7 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?
You should be a teacher Damon
To determine the lowest frequency photon that can accomplish ionization of an atom, we need to use the relationship between energy, frequency, and wavelength of light.
The energy (E) of a photon is given by the equation: E = hf
where h is Planck's constant (approximately 6.63 x 10^-34 J·s) and f is the frequency of the photon.
Given that 14.7 eV is needed to ionize the atom, we need to convert it to joules to make the calculation easier. Since 1 eV is equivalent to 1.6 x 10^-19 J, we have:
Ionization energy (E) = 14.7 eV * 1.6 x 10^-19 J/eV = 2.352 x 10^-18 J
Now, we can rearrange the energy equation to solve for frequency:
f = E / h
Plugging in the values, we get:
f = 2.352 x 10^-18 J / 6.63 x 10^-34 J·s = 3.55 x 10^15 Hz
To find the corresponding wavelength (λ), we can use the formula: λ = c / f
where c is the speed of light, approximately 3 x 10^8 m/s.
Plugging in the values, we get:
λ = 3 x 10^8 m/s / 3.55 x 10^15 Hz = 8.45 x 10^-8 m or 84.5 nm
Therefore, the lowest frequency photon required to ionize the atom has a frequency of 3.55 x 10^15 Hz and a corresponding wavelength of 84.5 nm.
h f = 14.7 eV
lambda = c T = c/f
c about 3*10^8 m/s