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In \(n_{R}\) and \(n_{V}\) denote the number of photons emitted by a red bulb and violet bulb of equal power in a given time, then (a) \(n_{R}=n_{V}\) (b) \(n_{R}>n_{V}\) (c) \(n_{R}

Short Answer

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(b) \(n_R > n_V\)

Step by step solution

01

Understanding the Relationship Between Frequency and Energy

Each photon emitted from a light source carries a specific amount of energy, which is related to its frequency by the equation \(E = h u\), where \(E\) is the energy, \(h\) is Planck's constant, and \(u\) is the frequency of the light.
02

Relating Frequency to Color Wavelengths

The frequency \(u\) of red light is lower than that of violet light owing to their respective positions in the visible spectrum, where violet light has a higher frequency (and shorter wavelength) than red light.
03

Calculating Photon Numbers for Equal Power Sources

For bulbs of equal power, the power \(P\) is distributed equally across photons emitted. If the violet bulb has higher energy photons (due to higher frequency), fewer photons \(n_V\) can be emitted than in the case of red light \(n_R\), where each photon carries less energy.
04

Conclusion of Photon Emission Comparison

Therefore, when both bulbs have the same power, more total photons are emitted by the red bulb than by the violet bulb, as per \(n_R > n_V\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frequency and Energy Relationship
The frequency of a light wave is a fundamental property that dictates its energy and, consequently, its color in the visible spectrum. According to the equation \( E = h u \), where \(E\) represents energy, \(h\) is Planck's constant, and \(u\) is the frequency, we can see a direct relationship between the frequency and the energy of a photon. This means:
  • The higher the frequency, the greater the energy of the photon.
  • Conversely, a lower frequency corresponds to a lower energy photon.
This relationship is crucial for understanding why different colors of light carry different amounts of energy. Violet light, residing at the higher frequency end of the visible spectrum, naturally has more energy per photon than red light, which is found at the lower frequency end.
Light Spectrum and Color Wavelengths
Colors of the visible spectrum are unique because they each have different wavelength and frequency characteristics. The visible spectrum ranges from violet to red, and each color within this range can be distinguished by its specific wavelength and frequency. Red light has a longer wavelength, which correlates with a lower frequency and thus lower energy. On the other hand:
  • Violet light has a shorter wavelength.
  • This equates to a higher frequency and greater energy per photon.
Because wavelength and frequency are inversely related, these characteristics together determine where a color appears in the spectrum, as well as its energy content. For students trying to understand why certain light behaves differently, noticing these wavelength differences is key in observing their practical effects.
Photons and Power Distribution
When considering light bulbs of equal power, the number of photons emitted depends on their individual energies, which vary based on the frequency. Power, which is the rate of energy used or transferred, remains constant for both the red and violet bulbs. Given their power is equal:
  • Fewer violet photons will be emitted due to their higher energy.
  • The red bulb can emit more photons because each one carries less energy.
This means \(n_R > n_V\) because more red photons fill the same energy budget. Therefore, when comparing red and violet light sources of the same power, red light will appear to have a broader distribution of photons making it more numerous than violet light.

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Most popular questions from this chapter

A source of light is placed at a distance of \(1 \mathrm{~m}\) from a photocell and cut-off potential is found to be \(V_{o}\). If the distance is doubled, the cut-off potential will be (a) \(2 V_{o}\) (b) \(V_{o}^{\prime 2}\) (c) \(V_{0}\) (d) \(V_{o} / 4\)

The frequency and the intensity of a beam of light falling on the surface of a photoelectric material are increased by a factor of two. This will (a) increase the maximum kinetic energy of the photoelectrons, as well as photoelectric current by a factor of two (b) increase the maximum kinetic energy of the photoelectrons and would increase the photoelectric current by a factor of two (c) increase the maximum kinetic energy of the photoelectrons by a factor of two and will have no effect on the magnitude of the photoelectric current produced (d) not produce any effect on the kinetic energy of the emitted electrons but will increase the photoelectric current by a factor of two

Which phenomenon best supports the theory that matter has a wave nature? (a) Electron momentum (b) Electron diffraction (c) Photon momentum (d) Photon diffraction

\(X\) -rays are not used for radar purpose because (a) they are not reflected by the target (b) they are not electromagnetic waves (c) they are completely absorbed by air (d) they sometimes damage the target

\(X\) -rays are known to be electromagnetic radiations. Therefore, an \(X\) -ray photon has (a) electric charge (b) magnetic moment (c) both electric charge and magnetic moment (d) neither electric charge nor magnetic moment

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