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Chapter 17: Problem 2343

An electric bulb of \(100 \mathrm{w}\) converts \(3 \%\) of electrical energy into light energy. If the wavelength of light emitted is \(6625 \AA\), the number of photons emitted is \(1 \mathrm{~s}\) is \(\ldots \ldots\) $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)$ (A) \(10^{17}\) (B) \(10^{19}\) (C) \(10^{21}\) (D) \(10^{15}\)

Short Answer

Expert verified
The number of photons emitted in 1 second is \(10^{19}\). The correct answer is (B) \(10^{19}\).

Step by step solution

01

Calculate the light energy emitted by the bulb in 1 second

Given that the bulb converts 3% of electrical energy into light energy, the light energy emitted by the bulb can be calculated as follows: Light energy = (3 / 100) * Power of the bulb * Time Light energy = (3 / 100) * 100 W * 1 s Light energy = 3 J
02

Find the energy of a single photon

The energy of a single photon can be calculated using the Planck's equation, which is given by: Energy of a photon (E) = Planck's constant (h) * Speed of light (c) / Wavelength (lambda) Given the wavelength (lambda) = 6625 Å = 6625 * 10^{-10} m and Planck's constant (h) = 6.625 * 10^{-34} Js, we can substitute the values in the equation: E = (6.625 * 10^{-34} Js) * (3 * 10^8 m/s) / (6625 * 10^{-10} m) E ≈ 3 * 10^{-19} J
03

Calculate the number of photons emitted in 1 second

We now have the total light energy emitted by the bulb (3 J) and the energy of a single photon (3 * 10^{-19} J). We can calculate the number of photons emitted in 1 second by dividing the total light energy by the energy of a single photon: Number of photons = Light energy / Energy of a single photon Number of photons = (3 J) / (3 * 10^{-19} J) Number of photons = 10^{19} The correct answer is (B) 10^{19}.

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Most popular questions from this chapter

Consider the radius of a nucleus to be \(10^{-15} \mathrm{~m} .\) If an electron is assumed to be in such nucleus, what ill be its energy? $\left(\mathrm{me}=9.1 \times 10^{-31} \mathrm{~kg} \cdot \mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)$ (A) \(5.59 \times 10^{3} \mathrm{MeV}\) (B) \(9.55 \times 10^{3} \mathrm{MeV}\) (C) \(5.95 \times 10^{3} \mathrm{MeV}\) (D) \(7.45 \times 10^{3} \mathrm{MeV}\)

De-Broglie wavelength of particle moving at a (1/4) th of speed of light having rest mass \(\mathrm{m}_{0}\) is \(\ldots \ldots \ldots\) (A) $\left\\{(3.87 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (B) $\left\\{(4.92 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (C) $\left\\{(7.57 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (D) $\left\\{(9.46 \mathrm{~h}) /\left(\mathrm{m}_{\circ} \mathrm{C}\right)\right\\}$

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) \(\left\\{\left(\mathrm{f}_{2}-\mathrm{f}_{1}\right) / \mathrm{K}\right\\}\)

Work functions for tungsten and sodium are \(4.5 \mathrm{eV}\) and $2.3 \mathrm{eV}\( respectively. If threshold wavelength of sodium is \)5460 \AA$, threshold wavelength for tungsten will be \(\ldots \ldots . . \AA\) (A) 528 (B) 10683 (C) 2791 (D) 5893

Select the correct statement from the following (A) Radiation and matter (particles) may not exhibit both the wave nature and particle nature simultaneously at the some moment. (B) At some moment electromagnetic waves get divided in to small pieces named particles (C) In a given circumstance a particle at one moment behaves like particle and the at the next moment as wave and so on (D) Each microscopic particle is enveloped by a wave
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