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Chapter 17: Problem 2417

A \(100 \mathrm{~W}\) bulb radiates energy at rate of $100 \mathrm{~J} / \mathrm{s}\(. If the light emitted has wavelength of \)525 \mathrm{~nm}$. How many photons are emitted per second? (A) \(1.51 \times 10^{20}\) (B) \(1.51 \times 10^{22}\) (C) \(2.64 \times 10^{20}\) (D) \(4.5 \times 10^{19}\)

Short Answer

Expert verified
The number of photons emitted per second by a $100 \mathrm{~W}$ light bulb with a wavelength of $525 \mathrm{~nm}$ is approximately \(2.64 \times 10^{20}\) photons per second.

Step by step solution

01

Write the formulas

We will use Planck's equation to find the energy per photon and then divide the power by the energy per photon to find the number of photons emitted per second. Planck's equation: E = (h × c) / λ Number of photons = Power / (Energy per photon)
02

Set up the constants

Planck's constant (h) = 6.626 x 10^-34 Js (joule-seconds) Speed of light (c) = 3.0 x 10^8 m/s (meters per second) Wavelength (λ) = 525 nm = 525 x 10^-9 m (converted to meters) Power = 100 J/s
03

Calculate the energy per photon

Using Planck's equation, we will calculate the energy per photon: E = (6.626 x 10^-34 Js) x (3.0 x 10^8 m/s) / (525 x 10^-9 m)
04

Solve for energy per photon

After solving the previous equation, we get: E ≈ 3.78 x 10^-19 J (joules)
05

Calculate the number of photons emitted per second

Now that we have found the energy per photon, we can calculate the number of photons emitted per second using our second formula: Number of photons = (100 J/s) / (3.78 x 10^-19 J)
06

Solve for the number of photons

After solving the equation, we get: Number of photons ≈ 2.64 x 10^20 photons per second
07

Choose the correct answer

The number of photons emitted per second is 2.64 x 10^20, which corresponds to option (C). Therefore, the correct answer is: (C) 2.64 x 10^20

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Most popular questions from this chapter

\(11 \times 10^{11}\) 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) \(2186 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\)

Frequency of incident light on body is \(\mathrm{f}\). Threshold frequency of body is \(f_{0}\). Maximum velocity of electron \(=\ldots \ldots \ldots\).. where \(m\) is mass of electron. (A) $\left[\left\\{2 \mathrm{~h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\right\\} / \mathrm{m}\right]^{(1 / 2)}$ (B) $\left[\left\\{2 \mathrm{~h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\right\\} / \mathrm{m}\right]$ (C) \([2 \mathrm{hf} / \mathrm{m}]^{(1 / 2)}\) (D) \(\mathrm{h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\)

Photoelectric effect is obtained on metal surface for a light having frequencies \(\mathrm{f}_{1} \& \mathrm{f}_{2}\) where \(\mathrm{f}_{1}>\mathrm{f}_{2}\). If ratio of maximum kinetic energy of emitted photo electrons is \(1: \mathrm{K}\), so threshold frequency for metal surface is \(\ldots \ldots \ldots \ldots\) (A) $\left\\{\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right) /(\mathrm{K}-1)\right\\}$ (B) $\left\\{\left(\mathrm{Kf}_{1}-\mathrm{f}_{2}\right) /(\mathrm{K}-1)\right\\}$ (C) $\left\\{\left(\mathrm{K} \mathrm{f}_{2}-\mathrm{f}_{1}\right) /(\mathrm{K}-1)\right\\}$ (D) \(\left\\{\left(\mathrm{f}_{2}-\mathrm{f}_{1}\right) / \mathrm{K}\right\\}\)

A proton falls freely under gravity of Earth. Its de Broglie wavelength after \(10 \mathrm{~s}\) of its motion is \(\ldots \ldots \ldots\). Neglect the forces other than gravitational force. $\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right), \mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.625 \times 10^{-34} \mathrm{~J}_{. \mathrm{s}}\right]$ (A) \(3.96 \AA\) (B) \(39.6 \AA\) (C) \(6.93 \AA\) (D) \(69.3 \AA\)

The work function of metal is \(5.3 \mathrm{eV}\). What is threshold frequency? (A) \(3.1 \times 10^{15} \mathrm{~Hz}\) (B) \(3.1 \times 10^{45} \mathrm{~Hz}\) (C) \(1.3 \times 10^{15} \mathrm{~Hz}\) (D) \(1.3 \times 10^{45} \mathrm{~Hz}\)
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