No idea how to do this. All I know is E=hf and found that a is 3.96 x 10^-19
Consider a sidewalk section that has an area of 1 m^2
and assume the sun is directly overhead.
Also, assuming the sun is beaming primarily green light, which has a wavelength of 500 nm. (a)
What is the energy of one sunlight photon? (b) Using the fact that the power of the sunlight on
this square meter is about 1367 W, how many photons hit the sidewalk section in one second? (c)
What is the momentum of one sunlight photon? (d) What is the momentum of all photons that hit
this sidewalk section in one second? (e) Now suppose an ant falls on the sidewalk. A typical ant
weighs 4x10^-6 kg and falls at 1.8 m/s. What is the ant’s momentum? (f) Which exerts a greater momentum, the ant or 1 second of sunlight? (g) Which would exert a greater momentum if the sidewalk were silvered, like a mirror
To find the answers to these questions, we need to use the formulas and information provided. Let's go step by step:
(a) The energy of one sunlight photon can be calculated using the equation E = hf, where E is the energy, h is Planck's constant (approximately 6.626 x 10^-34 J·s), and f is the frequency. Since we know the speed of light (c) is approximately 3.00 x 10^8 m/s, we can calculate the frequency using the formula f = c/λ, where λ is the wavelength. Given that the wavelength is 500 nm (or 500 x 10^-9 m), we have:
f = (3.00 x 10^8 m/s) / (500 x 10^-9 m) = 6.00 x 10^14 Hz.
Now, using the equation E = hf:
E = (6.626 x 10^-34 J·s) x (6.00 x 10^14 Hz) = 3.98 x 10^-19 J.
Therefore, the energy of one sunlight photon is approximately 3.98 x 10^-19 J.
(b) To determine how many photons hit the sidewalk section in one second, we need to use the power of sunlight on 1 square meter (1367 W) and the energy of one photon (3.98 x 10^-19 J). The power (P) is given by the formula P = E/t, where t is the time. Rearranging the formula, we get:
t = E / P.
t = (3.98 x 10^-19 J) / (1367 W) = 2.91 x 10^-22 s.
To find the number of photons, we divide the time (t) by the period (T) of one photon, where T = 1/f:
Number of photons = t / T = (2.91 x 10^-22 s) / (1 / 6.00 x 10^14 s^-1) = 1.75 x 10^-8 photons.
Therefore, approximately 1.75 x 10^-8 photons hit the sidewalk section in one second.
(c) The momentum (p) of one photon can be calculated using the equation p = E / c, where c is the speed of light:
p = (3.98 x 10^-19 J) / (3.00 x 10^8 m/s) = 1.33 x 10^-27 kg·m/s.
Therefore, the momentum of one sunlight photon is approximately 1.33 x 10^-27 kg·m/s.
(d) To find the momentum of all photons hitting the sidewalk section in one second, we multiply the momentum of one photon (1.33 x 10^-27 kg·m/s) by the number of photons (1.75 x 10^-8 photons):
Momentum = (1.33 x 10^-27 kg·m/s) x (1.75 x 10^-8 photons) = 2.33 x 10^-35 kg·m/s.
Therefore, the momentum of all photons that hit the sidewalk section in one second is approximately 2.33 x 10^-35 kg·m/s.
(e) The momentum of the ant can be calculated using the formula p = mv, where m is the mass and v is the velocity:
p = (4 x 10^-6 kg) x (1.8 m/s) = 7.2 x 10^-6 kg·m/s.
Therefore, the momentum of the ant is approximately 7.2 x 10^-6 kg·m/s.
(f) To compare the momentum of the ant and 1 second of sunlight, we can directly compare their magnitudes. Comparing the values, we find that the magnitude of the momentum of the ant (7.2 x 10^-6 kg·m/s) is greater than the magnitude of the momentum of 1 second of sunlight (2.33 x 10^-35 kg·m/s).
Therefore, the ant exerts a greater magnitude of momentum than 1 second of sunlight.
(g) If the sidewalk section were silvered like a mirror, it would reflect the sunlight, meaning there would be a change in momentum when the light is reflected. However, since the mirror does not change the magnitude of momentum, the momentum of the reflected light would still be the same as the incident sunlight.
Therefore, the momentum of the sunlight hitting the silvered sidewalk section would be the same as the momentum calculated in part (d) (approximately 2.33 x 10^-35 kg·m/s).