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Output power of He-Ne LASER of low energy is \(1.00 \mathrm{~mW}\). Wavelength of the light is \(632.8 \mathrm{~nm}\). What will be the number of photons emitted per second from this LASER? (A) \(8.31 \times 10^{15} \mathrm{~s}^{-1}\) (B) \(5.38 \times 10^{15} \mathrm{~s}^{-1}\) (C) \(1.83 \times 10^{15} \mathrm{~s}^{-1}\) (D) \(3.18 \times 10^{15} \mathrm{~s}^{-1}\)

Short Answer

Expert verified
The number of photons emitted per second from the He-Ne LASER is \(5.38 \times 10^{15} \,s^{-1}\).

Step by step solution

01

Convert the values to the correct units

First, we need to make sure all our values are in the correct units. We need to convert the LASER power to watts and the wavelength to meters. The output power of the He-Ne LASER is given as 1.00 mW (milliwatts). To convert it to watts, we multiply by a conversion factor: Power = \(1.00 \times 10^{-3} \,W\) The wavelength is given as 632.8 nm (nanometers). To convert it to meters, we also multiply by a conversion factor: Wavelength = \(632.8 \times 10^{-9} \,m\)
02

Calculate the energy of a single photon

We can now calculate the energy of a single photon using the energy of a photon formula: Energy = \(\frac{hc}{\lambda}\) Here, 'h' is Planck's constant: \(h = 6.626 \times 10^{-34} \, Js\), 'c' is the speed of light: \(c = 3 \times 10^8 \, \frac{m}{s}\), and '\(\lambda\)' is the wavelength of the light: \(\lambda = 632.8 \times 10^{-9} \,m\). Plugging the values into the formula, we get: Energy = \(\frac{(6.626 \times 10^{-34} \, Js) (3 \times 10^8 \, \frac{m}{s})}{632.8 \times 10^{-9} \,m}\) Energy = \(3.141 \times 10^{-19} \,J\)
03

Calculate the number of photons emitted per second

Now, we can find the number of photons emitted per second by dividing the LASER power (total energy emitted per second) by the energy of a single photon: Number of photons = \(\frac{LASER \, power}{Energy \, of \, a \, single \, photon}\) Number of photons = \(\frac{1.00 \times 10^{-3} \,W}{3.141 \times 10^{-19} \,J}\) Number of photons = \(5.38 \times 10^{15} \, photons/s\) The number of photons emitted per second from the He-Ne LASER is \(5.38 \times 10^{15} \,s^{-1}\). So the correct answer is (B).

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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) \(3.29 \times 10^{-14} \mathrm{~m}\)

Wavelength of light incident on a photo-sensitive surface is reduced form \(3500 \AA\) to \(290 \mathrm{~mm}\). The change in stopping potential is $\ldots \ldots . .\left(\mathrm{h}=6.625 \times 10^{-24} \mathrm{~J} . \mathrm{s}\right)$ (A) \(42.73 \times 10^{-2} \mathrm{~V}\) (B) \(27.34 \times 10^{-2} \mathrm{~V}\) (C) \(73.42 \times 10^{-2} \mathrm{~V}\) (D) \(43.27 \times 10^{-2} \mathrm{~V}\)

Read the paragraph carefully and select the proper choice from given multiple choices. According to Einstein when a photon of light of frequency for wavelength \(\lambda\) is incident on a photo sensitive metal surface of work function \(\Phi\). Where \(\Phi<\mathrm{hf}\) (here \(\mathrm{h}\) is Plank's constant) then the emission of photo-electrons place takes place. The maximum K.E. of emitted photo electrons is given by $\mathrm{K}_{\max }=\mathrm{hf}-\Phi .\( If the there hold frequency of metal is \)\mathrm{f}_{0}$ then \(\mathrm{hf}_{0}=\Phi\) (i) A metal of work function \(3.3 \mathrm{eV}\) is illuminated by light of wave length \(300 \mathrm{~nm}\). The maximum \(\mathrm{K} . \mathrm{E}>\) of photo- electrons is $=\ldots \ldots \ldots . \mathrm{eV} .\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{sec}\right)$ (A) \(0.825\) (B) \(0.413\) (C) \(1.32\) (D) \(1.65\) (ii) Stopping potential of emitted photo-electron is $=\ldots \ldots . \mathrm{V}$. (A) \(0.413\) (B) \(0.825\) (C) \(1.32\) (D) \(1.65\) (iii) The threshold frequency fo $=\ldots \ldots \ldots \times 10^{14} \mathrm{~Hz}$. (A) \(4.0\) (B) \(4.2\) (C) \(8.0\) (D) \(8.4\)

Matching type questions: (Match, Column-I and Column-II property) Column-I Column-II (I) Energy of photon of wavelength \(\lambda\) is (P) \((\mathrm{E} / \mathrm{p})\) (II) The de Broglie wavelength associated (Q) \(\left(\mathrm{hf} / \mathrm{c}^{2}\right)\) with particle of momentum \(\mathrm{P}\) is (II) Mass of photon in motion is (R) (hc \(/ \lambda\) ) (IV) The velocity of photon of energy (S) \((\mathrm{h} / \mathrm{p})\) \(\mathrm{E}\) and momentum \(\mathrm{P}\) is (A) I - P, II - Q. III - R, IV - S (B) $\mathrm{I}-\mathrm{R}, \mathrm{II}-\mathrm{S}, \mathrm{III}-\mathrm{Q}, \mathrm{IV}-\mathrm{P}$ (C) $\mathrm{I}-\mathrm{R}, \mathrm{II}-\mathrm{S}, \mathrm{III}-\mathrm{P}_{3} \mathrm{IV}-\mathrm{Q}$ (D) $\mathrm{I}-\mathrm{S}, \mathrm{II}-\mathrm{R}, \mathrm{III}-\mathrm{Q}, \mathrm{IV}-\mathrm{P}$

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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