Does a photon emitted by a higher-wattage red light bulb have more energy than a photon emitted by a lower-wattage red bulb
1 answer:
Answer:
The energy of these two photons would be the same as long as their frequencies are the same (same color, assuming that the two bulbs emit at only one wavelength.)
Explanation:
The energy of a photon is proportional to its frequency
. The constant of proportionality is Planck's Constant,
. This proportionality is known as the Planck-Einstein Relation.
.
The color of a beam of visible light depends on the frequency of the light. Assume that the two bulbs in this question each emits light of only one frequency (rather than a mix of light of different frequencies and colors.) Let and
denote the frequency of the light from each bulb.
If the color of the red light from the two bulbs is the same, those two bulbs must emit light at the same frequency: .
Thus, by the Planck-Einstein Relation, the energy of a photon from each bulb would also be the same:
.
Note that among these two bulbs, the brighter one appears brighter soley because it emits more photons per unit area in unit time. While the energy of each photon stays the same, the bulb releases more energy by emitting more of these photons.
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Answer:
Maximum height reached by the ball is 32 meters.
Explanation:
It is given that,
If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :
...........(1)
t is the time taken
s is the height attained as a function of time.
Maximum height achieved can be calculated as :
-16 t + 32 = 0
t = 2 seconds
Put the value of t in equation (1) as :
s = 32 meters
So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.
a) Total power output:
b) The relative percentage change of power output is 1.67%
c) The intensity of the radiation on Mars is
Explanation:
a)
The intensity of electromagnetic radiation is given by
where
P is the power output
A is the surface area considered
In this problem, we have
is the intensity of the solar radiation at the Earth
The area to be considered is area of a sphere of radius
(distance Earth-Sun)
Therefore
And now, using the first equation, we can find the total power output of the Sun:
b)
The energy of the solar radiation is directly proportional to its frequency, given the relationship
where E is the energy, h is the Planck's constant, f is the frequency.
Also, the power output of the Sun is directly proportional to the energy,
where t is the time.
This means that the power output is proportional to the frequency:
Here the frequency increases by 1 MHz: the original frequency was
so the relative percentage change in frequency is
And therefore, the power also increases by 1.67 %.
c)
In this second case, we have to calculate the new power output of the Sun:
Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is
Now we can find the radiation intensity with the equation
Where the area is
And substituting,
Learn more about electromagnetic radiation:
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