The maximum wavelength of light emitted from a blackbody that will emit electrons from a given piece of metal is 255nm. What is the fraction of total energy emitted from the black body that emits electrons?
I set up the integral: Int(8πhcλ^-5)/(e^(hc/λkT)-1)) from 0 to 255nm all over the same integral from 0 to infinity (planck's law) but I have NO IDEA how to solve it! THANKS
They are asking for the fraction of the blackbody radiation that has a wavelength less than 255 nm.
This can only be computed if you already know the TEMPERATURE of the blackbody.
Setting up the integral is the right way to go about answering the question, but you still need the temperature.
Here is how it is phased in the question:
(2) The orbiting space shuttle moves around the Earth well above 99% of the atmosphere, yet it still accumulates an electric charge on its skin due (in part) to the loss of electrons caused by the photoelectric effect from sunlight. Suppose the skin of the shuttle is coated with nickel for which the work function is φ = 4.87 eV at the temperatures encountered while in orbit. (A) What is the maximum wavelength of solar radiation that can result in electron emission from the shuttle’s skin? (B) What is the maximum fraction of the total power falling on the shuttle that could potentially produce photoelectrons?
I found part A to be 255nm. But how do you find the temperature from that information?
To solve the integral and find the fraction of total energy emitted from the blackbody that will emit electrons, you need to evaluate two separate integrals: one from 0 to 255 nm and another from 0 to infinity.
Let's start with the first integral, which represents the energy emitted only up to 255 nm. We'll call this integral A:
A = ∫[0 to 255 nm] (8πhcλ^(-5))/(e^(hc/λkT)-1) dλ
To solve this integral, you can use numerical methods or specialized software like Mathematica. However, in this case, we can approximate the integral by discretizing the wavelength into small intervals and using the trapezoidal rule for numerical integration.
1. Divide the wavelength range [0, 255 nm] into N small intervals.
2. Calculate the width of each interval: Δλ = (255 nm - 0) / N.
3. Evaluate the function at each interval's midpoint: λ_i = Δλ/2, λ_i+1 = λ_i + Δλ, ..., λ_N-1 = λ_N - Δλ/2.
4. Substitute these values into the integrand and sum them up, multiplying by Δλ/2:
A ≈ Δλ/2 * ∑[i=0 to N-1] ((8πhcλ_i^(-5))/(e^(hc/λ_i*k*T)-1) + (8πhcλ_i+1^(-5))/(e^(hc/λ_i+1*k*T)-1)).
This approximation will give you a numerical value for A, representing the energy emitted up to 255 nm.
Next, we need to evaluate the integral B, which represents the total energy emitted from the blackbody, extending to infinity:
B = ∫[0 to ∞] (8πhcλ^(-5))/(e^(hc/λkT)-1) dλ.
The integral B is known as the Stefan-Boltzmann law and represents the total energy radiation from a blackbody. Its value has already been derived and is approximately equal to σT^4 (where σ is the Stefan-Boltzmann constant). Therefore, B = σT^4.
Finally, to find the fraction of total energy emitted from the blackbody that will emit electrons, you can divide integral A by integral B:
Fraction = A / B
This fraction will give you the desired result, indicating the proportion of energy emitted up to 255 nm compared to the total energy emitted by the blackbody.