What’s the answer please?
1 answer:
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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.
Answer:
E=
Explanation:
We are given that
Charge on ring= Q
Radius of ring=a
We have to find the magnitude of electric filed on the axis at distance a from the ring's center.
We know that the electric field at distance x from the center of ring of radius R is given by
Substitute x=a and R=a
Then, we get
Where K=
Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=