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Chapter 27: Problem 62

An owl has good night vision because its eyes can detect a light intensity as faint as \(5.0 \times 10^{-13} \mathrm{W} / \mathrm{m}^{2} .\) What is the minimum number of photons per second that an owl eye can detect if its pupil has a diameter of \(8.5 \mathrm{mm}\) and the light has a wavelength of $510 \mathrm{nm} ?$

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Step by step solution

01

Calculate the area of the owl's pupil

First, let's find the area of the owl's pupil using the given diameter. The formula for the area of a circle is: \(A = \pi r^{2}\), where A is the area and r is the radius. We need to convert the diameter to radius by dividing by 2. So, \(r = \frac{8.5 \mathrm{mm}}{2} = 4.25 \mathrm{mm}\). Now we need to convert the radius to meters: \(r = 4.25 \times 10^{-3} \mathrm{m}\). Now, we can calculate the area: \(A = \pi (4.25 \times 10^{-3})^{2} \mathrm{m^2}\).
02

Find the energy of each photon

To find the energy of each photon, we can use the formula: \(E = \frac{hc}{\lambda}\), where E is the energy, h is the Planck's constant (\(6.63 \times 10^{-34} \mathrm{Js}\)), c is the speed of light (\(3 \times 10^8 \mathrm{m/s}\)), and λ is the wavelength. We are given λ = \(510 \mathrm{nm}\). First, we need to convert it to meters: \(\lambda = 510 \times 10^{-9} \mathrm{m}\). Now, we can calculate the energy of each photon: \(E = \frac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{m/s}}{510 \times 10^{-9} \mathrm{m}}\).
03

Calculate the number of photons detected per second

Now, we need to calculate the number of photons detected per second. We can do this by dividing the light intensity by the energy per photon and then multiplying by the area of the pupil. The formula is: \(N = \frac{I \times A}{E}\), where N is the number of photons per second, I is the light intensity, A is the area of the pupil, and E is the energy per photon. We have \(I = 5.0 \times 10^{-13} \mathrm{W/m^2}\), A from Step 1 and E from Step 2. Now, we can calculate the number of photons detected per second: \(N = \frac{5.0 \times 10^{-13} \mathrm{W/m^2} \times \pi (4.25 \times 10^{-3})^{2} \mathrm{m^2}}{E}\). To find the minimum number of photons, round up the result to the nearest whole number.

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Most popular questions from this chapter

(a) What is the difference in radius between the \(n=1\) state and the \(n=2\) state for hydrogen? (b) What is the difference in radius between the \(n=100\) state and the \(n=101\) state for hydrogen? How do the neighboring orbital separations compare for large and small \(n\) values?
Follow the steps outlined in this problem to estimate the time lag (predicted classically but not observed experimentally) in the photoelectric effect. Let the intensity of the incident radiation be $0.01 \mathrm{W} / \mathrm{m}^{2} .\( (a) If the area of the atom is \)(0.1 \mathrm{nm})^{2},$ find the energy per second falling on the atom. (b) If the work function is 2.0 eV, how long would it take (classically) for enough energy to fall on this area to liberate one photoelectron? (c) Explain briefly, using the photon model, why this time lag is not observed.
In a color TV tube, electrons are accelerated through a potential difference of \(20.0 \mathrm{kV} .\) Some of the electrons strike the metal mask (instead of the phosphor dots behind holes in the mask), causing x-rays to be emitted. What is the smallest wavelength of the \(x\) -rays emitted?
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