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Chapter 36: Problem 40

If \(10 \%\) of the energy of a \(100 .\) W incandescent bulb is in the form of visible light having an average wavelength of \(600 .\) nm, how many photons are emitted per second from the light bulb?

Short Answer

Expert verified
Answer: Approximately 3.01 × 10^{19} photons are emitted per second.

Step by step solution

01

Calculate Energy Emitted per Second in Visible Light

To find the energy emitted per second in visible light, we first need to calculate the total power in visible light. We have the percentage of power in visible light and the total power of the bulb, and can use this to calculate the power in visible light. Power in visible light = Total power of bulb * (Percentage of power in visible light / 100) Power in visible light = 100 W * (10% / 100) = 10 W Next, we have to determine the energy emitted per second in visible light. Energy Emitted per Second in Visible Light = Power in Visible Light Energy Emitted per Second in Visible Light = 10 J/s
02

Calculate Energy per Photon

Now, we will calculate the energy per photon using the formula: Energy per Photon = (Planck's constant * Speed of light) / Wavelength where Planck's constant h = 6.63 × 10^{-34} Js, the speed of light c = 3 × 10^8 m/s, and the wavelength λ = 600 nm = 600 × 10^{-9} m. Energy per Photon = (6.63 × 10^{-34} Js * 3 × 10^8 m/s) / (600 × 10^{-9} m) Energy per Photon = 3.315 × 10^{-19} J
03

Calculate the Number of Photons Emitted per Second

Now that we have the energy emitted per second in visible light and the energy per photon, we can calculate the number of photons emitted per second using the formula: Number of Photons per Second = Energy Emitted per Second in Visible Light / Energy per Photon Number of Photons per Second = 10 J/s / 3.315 × 10^{-19} J Number of Photons per Second ≈ 3.01 × 10^{19} So, approximately 3.01 × 10^{19} photons are emitted per second from the 100 W incandescent bulb.

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Key Concepts

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

Planck's Constant
Planck's Constant, denoted by the symbol \( h \), is a fundamental constant in physics. It plays a crucial role in the field of quantum mechanics by linking the energy of a photon to the frequency of electromagnetic waves. Specifically, Planck's Constant helps explain how energy is quantized in tiny, discrete packets known as quanta. In essence, it shows that energy is not continuous, but rather comes in small chunks or packets, which we now understand as photons.- The value of Planck's Constant is approximately \( 6.63 \times 10^{-34} \) Joules per second (\( Js \)).- It is fundamental to the equations relating to wave-particle duality, where particles like electrons can exhibit both wave-like and particle-like properties.When calculating the energy of a photon, we often use the formula: \[ E = h \cdot f \]where \( E \) is the energy in Joules, \( h \) is Planck's constant, and \( f \) is the frequency of the wave in Hertz.
Energy per Photon
The energy per photon is a measure of the energy carried by a single photon, which is a particle of light. To find this energy, one can use the relationship between a photon's energy, its wavelength, and two constants—Planck's constant and the speed of light. The energy per photon can be calculated using the formula:\[ E = \frac{h \cdot c}{\lambda} \]where:
  • \( E \) is the energy per photon in Joules.
  • \( h \) is Planck's constant, \( 6.63 \times 10^{-34} \) Js.
  • \( c \) is the speed of light, approximately \( 3 \times 10^8 \) meters per second.
  • \( \lambda \) is the wavelength of the light in meters.
In scenarios such as an incandescent bulb emitting visible light, calculating the energy per photon allows us to determine how much energy each photon of light comprises. This understanding is essential for figuring out how many photons are emitted per second when light is produced, given a certain amount of energy is emitted.
Visible Light
Visible light is the range of electromagnetic radiation that the human eye can detect. It encompasses a spectrum of wavelengths, typically from about 380 nanometers (nm) to 750 nanometers. This range is responsible for the spectrum of colors that we can see, from violet on the lower end to red on the upper end. Understanding visible light is important when studying phenomena like photon emission, as in the case of a light bulb, where photons are the carriers of light energy that becomes visible to us. - **Wavelengths in Visible Light:** Different wavelengths within the visible spectrum correspond to different colors, with shorter wavelengths like violet having more energy per photon compared to longer wavelengths like red. - **Energy Distribution:** In practical scenarios, like a light bulb, not all emitted light is visible; for instance, only about 10% of an incandescent bulb's energy might be in the visible spectrum, with the rest in infrared. Visible light is just one part of the electromagnetic spectrum, but it holds special significance for us due to its visibility and role in the perception of the world around us.

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Most popular questions from this chapter

Consider the following mechanism, which results in the formation of product \(P:\) \\[ \begin{array}{l} \mathrm{A} \stackrel{k_{1}}{\rightleftharpoons_{k-1}} \mathrm{B} \frac{k_{2}}{\rightleftharpoons_{-2}} \mathrm{C} \\ \mathrm{B} \stackrel{k_{3}}{\rightarrow} \mathrm{P} \end{array} \\] If only the species \(A\) is present at \(t=0,\) what is the expression for the concentration of \(\mathrm{P}\) as a function of time? You can apply the preequilibrium approximation in deriving your answer.

The Rice-Herzfeld mechanism for the thermal decomposition of acetaldehyde \(\left(\mathrm{CH}_{3} \mathrm{CO}(g)\right)\) is \\[ \begin{array}{l} \mathrm{CH}_{3} \mathrm{CHO}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{CH}_{3} \cdot(g)+\mathrm{CHO} \cdot(g) \\ \mathrm{CH}_{3} \cdot(g)+\mathrm{CH}_{3} \mathrm{CHO}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CH}_{4}(g)+\mathrm{CH}_{2} \mathrm{CHO} \cdot(g) \\ \mathrm{CH}_{2} \mathrm{CHO} \cdot(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{CO}(g)+\mathrm{CH}_{3} \cdot(g) \\ \mathrm{CH}_{3} \cdot(g)+\mathrm{CH}_{3} \cdot(g) \stackrel{k_{4}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g) \end{array} \\] Using the steady-state approximation, determine the rate of methane \(\left(\mathrm{CH}_{4}(g)\right)\) formation.

The quantum yield for \(\mathrm{CO}(g)\) production in the photolysis of gaseous acetone is unity for wavelengths between 250 and \(320 \mathrm{nm} .\) After 20.0 min of irradiation at \(313 \mathrm{nm}\) \(18.4 \mathrm{cm}^{3}\) of \(\mathrm{CO}(g)\) (measured at \(1008 \mathrm{Pa}\) and \(22^{\circ} \mathrm{C}\) ) is produced. Calculate the number of photons absorbed and the absorbed intensity in \(\mathrm{J} \mathrm{s}^{-1}\)

The adsorption of ethyl chloride on a sample of charcoal at \(0^{\circ} \mathrm{C}\) measured at several different pressures is as follows: $$\begin{array}{rc} \mathbf{P}_{C_{2} H_{5} C l}(\text { Torr }) & V_{\text {ads}}(\mathbf{m L}) \\\ \hline 20 & 3.0 \\ 50 & 3.8 \\ 100 & 4.3 \\ 200 & 4.7 \\ 300 & 4.8 \end{array}$$ Using the Langmuir isotherm, determine the fractional coverage at each pressure and \(V_{m}\)

Reciprocal plots provide a relatively straightforward way to determine if an enzyme demonstrates Michaelis-Menten kinetics and to determine the corresponding kinetic parameters. However, the slope determined from these plots can require significant extrapolation to regions corresponding to low substrate concentrations. An alternative to the reciprocal plot is the Eadie- Hofstee plot in which the reaction rate is plotted versus the rate divided by the substrate concentration and the data are fit to a straight line. a. Beginning with the general expression for the reaction rate given by the Michaelis-Menten mechanism: \\[ R_{0}=\frac{R_{\max }[\mathrm{S}]_{0}}{[\mathrm{S}]_{0}+K_{m}} \\] rearrange this equation to construct the following expression, which is the basis for the Eadie-Hofstee plot: \\[ R_{0}=R_{\max }-K_{m}\left(\frac{R_{0}}{[S]_{0}}\right) \\] b. Using an Eadie-Hofstee plot, determine \(R_{\max }\) and \(K_{m}\) for hydrolysis of sugar by the enzyme invertase using the following data: $$\begin{array}{cc} \text { [Sucrose ] }_{\mathbf{0}}(\mathbf{M}) & \mathbf{R}_{\mathbf{0}}\left(\mathbf{M} \mathbf{~ s}^{-\mathbf{1}}\right) \\ \hline 0.029 & 0.182 \\ 0.059 & 0.266 \\ 0.088 & 0.310 \\ 0.117 & 0.330 \\ 0.175 & 0.362 \\ 0.234 & 0.361 \end{array}$$
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