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Chapter 17: Problem 2337

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

Work function of metal is \(2 \mathrm{eV}\). Light of intensity $10^{-5} \mathrm{Wm}^{-2}\( is incident on \)2 \mathrm{~cm}^{2}\( area of it. If \)10^{17}$ electrons of these metals absorb the light, in how much time does the photo electric effectc start? Consider the waveform of incident light. (A) \(1.4 \times 10^{7} \mathrm{sec}\) (B) \(1.5 \times 10^{7} \mathrm{sec}\) (C) \(1.6 \times 10^{7} \mathrm{sec}\) (D) \(1.7 \times 10^{7} \mathrm{sec}\)

Which of the following phenomenon can not be explained by quantum theory of light? (A) Emission of radiation from black body (B) Photo electric effect (C) Polarization (D) Crompton effect

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) \(\mathrm{h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\)

Matching type questions: (Match, Column-I and Column-II property) Column-I Column-II (A) Planck's theory of quantum (p) Light energy \(=\mathrm{hv}\) (B) Einstein's theory of quanta (q) Angular momentum of electron in an orbit. (C) Bohr's stationary orbit (r) Oscillator energies (D) D-Broglie waves (s) Electron microscope (A) \((\mathrm{A}-\mathrm{p}),(\mathrm{b}-\mathrm{q}),(\mathrm{C}-\mathrm{r}),(\mathrm{D}-\mathrm{s})\) (B) \((\mathrm{A}-\mathrm{q}),(\mathrm{B}-\mathrm{r}),(\mathrm{C}-\mathrm{s}),(\mathrm{D}-\mathrm{p})\) (C) \((\mathrm{A}-\mathrm{r}),(\mathrm{B}-\mathrm{p}),(\mathrm{C}-\mathrm{q}),(\mathrm{D}-\mathrm{s})\) (D) \((\mathrm{A}-\mathrm{r}),(\mathrm{B}-\mathrm{p}),(\mathrm{C}-\mathrm{s}),(\mathrm{D}-\mathrm{q})\)

The de-Broglie wavelength of a proton and \(\alpha\) - particle is same. The ratio of their velocities will be.......... ( \(\alpha\) particle is the He- nucleus, having two protons and two neutrons. Thus, its mass \(\mathrm{M}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\) where \(\mathrm{m}_{\mathrm{p}}\) is the mass of the proton.) (A) \(1: 4\) (B) \(1: 2\) (C) \(2: 1\) (D) \(4: 1\)
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