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Chapter 17: Problem 2318

When electric bulb having \(100 \mathrm{~W}\) efficiency emits photon having wavelength \(540 \mathrm{~nm}\) every second, numbers of photons will be $\ldots \ldots\left(\mathrm{h}=6 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right)$ (A) 100 (B) 1000 (C) \(3 \times 10^{20}\) (D) \(3 \times 10^{18}\)

Short Answer

Expert verified
The number of photons emitted per second by the electric bulb is \(3 \times 10^{20}\).

Step by step solution

01

Write down the energy of a single photon formula

To find the energy of a single photon, we use the following formula: \[E = \dfrac{hc}{\lambda}\] where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.
02

Calculate the energy of a single photon

Plug the given values into the formula and solve for the energy of a single photon: \[E = \dfrac{(6 \times 10^{-34} \, J \cdot s)(3 \times 10^8 \, ms^{-1})}{540 \times 10^{-9} \, m}\] \[E = 3.33 \times 10^{-19} \, J\]
03

Write down the formula to find the number of photons emitted per second

To determine the number of photons emitted per second, we need to divide the total energy emitted per second (Power of the bulb) by the energy of a single photon: \[N=\dfrac{P}{E}\] where N is the number of photons emitted per second, P is the power of the bulb (efficiency), and E is the energy of a single photon.
04

Calculate the number of photons emitted per second

Plug the values calculated in the previous steps and the given power of the bulb (100 W) into the formula and solve for the number of photons per second: \[N=\dfrac{100 \, J/s}{3.33 \times 10^{-19} \, J}\] \[N = 3 \times 10^{20}\] So, the number of photons emitted per second is \(3 \times 10^{20}\). The correct option is (C).

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Most popular questions from this chapter

An electron moving with velocity \(0.6 \mathrm{c}\), then de-brogly wavelength associated with is \(\ldots \ldots \ldots\) (rest mars of electron, \(\mathrm{m}_{0}=9.1 \times 10^{-31}(\mathrm{k} / \mathrm{s})\) \(\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\) (A) \(3.24 \times 10^{-12} \mathrm{~m}\) (B) \(32.4 \times 10^{-12} \mathrm{~m}\) (C) \(320 \times 10^{-12} \mathrm{~m}\) (D) \(3.29 \times 10^{-14} \mathrm{~m}\)

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A photon, an electron and a uranium nucleus all have same wavelength. The one with the most energy \(\ldots \ldots \ldots \ldots\) (A) is the photon (B) is the electron (C) is the uranium nucleus (D) depends upon the wavelength and properties of the particle.

\(11 \times 10^{11}\) Photons are incident on a surface in \(10 \mathrm{~s}\). These photons correspond to a wavelength of \(10 \AA\). If the surface area of the given surface is \(0.01 \mathrm{~m}^{2}\), the intensity of given radiations is \(\ldots \ldots\) $\left\\{\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(21.86 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) (B) \(2.186 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) (C) \(218.6 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) (D) \(2186 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\)
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