I have a metal sheet of indium, and when a uv light with a wavelength of 265 nm is shone on it the surface ejects electrons that have a kinetic energy of 8.31x10^-20 J per ejected electron. I'm asked to find the threshold frequency, I need to know how to find it.
To find the threshold frequency, we need to use the relationship between the energy of a photon and its frequency. The equation that relates these two values is given by:
E = h * f
where E is the energy of a photon, h is Planck's constant (6.626 x 10^-34 J*s), and f is the frequency of the photon.
In this case, we know the kinetic energy (8.31 x 10^-20 J) of the ejected electron per photon. We can use this information to find the energy of a photon.
The kinetic energy of the ejected electron is equal to the energy of the incident photon minus the work function (ϕ) of the metal:
Kinetic Energy = Energy of the Photon - Work Function
Since the incident light is UV light, we can assume that this energy is equivalent to the energy of a single photon. Thus, we can rewrite the equation as:
8.31 x 10^-20 J = h * f - ϕ
We re-arrange the equation to solve for the frequency (f):
h * f = 8.31 x 10^-20 J + ϕ
f = (8.31 x 10^-20 J + ϕ) / h
To find the threshold frequency, we need to determine the work function of the metal (ϕ). The work function is the minimum amount of energy required to remove an electron from the metal surface. In this case, it represents the kinetic energy of the ejected electron.
Therefore, to find the threshold frequency, we need to calculate:
f = (8.31 x 10^-20 J + 8.31 x 10^-20 J) / (6.626 x 10^-34 J*s)
f = 1.26 x 10^14 Hz
So, the threshold frequency is approximately 1.26 x 10^14 Hz.