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 can use the equation that relates the kinetic energy of an ejected electron to the energy of a photon:
Kinetic energy = hν - φ
In this equation, h is the Planck's constant (6.626 x 10^-34 J·s), ν is the frequency of the photon, and φ is the work function of the material (which corresponds to the minimum energy needed to remove an electron from the material).
We know the kinetic energy of the ejected electron (8.31 x 10^-20 J), and we want to find the corresponding frequency (threshold frequency).
Step 1: Convert the kinetic energy to joules
Since the kinetic energy is given in joules, we can skip this step.
Step 2: Plug the given values into the equation
8.31 x 10^-20 J = hν - φ
Step 3: Rearrange the equation to isolate the frequency
hν = 8.31 x 10^-20 J + φ
Step 4: Subtract the work function from both sides
hν - φ = 8.31 x 10^-20 J
Step 5: Solve for the frequency
ν = (8.31 x 10^-20 J + φ) / h
Now, we need to find the work function of indium. The work function can be calculated using the equation:
φ = hc / λ
Where h is the Planck's constant, c is the speed of light (3 x 10^8 m/s), and λ is the wavelength of the UV light.
Step 6: Convert the given UV light wavelength of 265 nm to meters
265 nm = 265 x 10^-9 m
Step 7: Plug the values into the equation
φ = (6.626 x 10^-34 J·s * 3 x 10^8 m/s) / (265 x 10^-9 m)
Step 9: Calculate the work function
φ = 7.52 x 10^-19 J
Step 10: Substitute the values back into the frequency equation
ν = (8.31 x 10^-20 J + 7.52 x 10^-19 J) / (6.626 x 10^-34 J·s)
Step 11: Calculate the threshold frequency
ν = 3.01 x 10^14 Hz
Therefore, the threshold frequency is approximately 3.01 x 10^14 Hz.
To find the threshold frequency, you can use the equation:
E = hf - φ
Where:
E is the kinetic energy of the ejected electron (8.31x10^-20 J),
h is the Planck's constant (6.62607015 × 10^-34 J·s),
f is the frequency of the light, and
φ is the work function of the material (the minimum energy required to remove an electron).
Since the kinetic energy can be written as:
E = (1/2)mv^2
Where:
m is the mass of the electron, and
v is the velocity of the electron.
Since the velocity of the electron can be written as:
v = λf
Where:
λ is the wavelength of the UV light (265 nm),
f is the frequency, and
c is the speed of light (3 x 10^8 m/s).
We can substitute the value of v into the equation for kinetic energy:
E = (1/2)m(λf)^2
Next, we can substitute the value of E into the equation E = hf - φ:
hf - φ = (1/2)m(λf)^2
Since we need to find the threshold frequency, we can assume that the ejected electron has no kinetic energy (E = 0), so the equation becomes:
0 = hf - φ
Therefore, the threshold frequency f₀ can be calculated as:
f₀ = φ / h
Note: The work function φ is the minimum energy required to remove an electron from the surface, and it can be derived from the given kinetic energy (8.31x10^-20 J) using φ = E.