A laser used in retinal surgery emits photons of wavelength \(632.8 \mathrm{~nm}\) in pulses of \(0.5 \mathrm{~W}\) that last \(2 \times 10^{-2} \mathrm{~s}\). The number of photons in each pulse is nearest to (A) \(10^{10}\) (B) \(10^{15}\) (C) \(10^{20}\) (D) \(10^{25}\) (E) \(10^{30}\)

Short Answer

Expert verified
Answer: (B) \(10^{15}\)

Step by step solution

01

Convert the wavelength to meters

Given the wavelength of the photons is \(632.8~\mathrm{nm}\), we need to convert this to meters so that we can use it in our calculations. There are \(1\times10^9\) nanometer in 1 meter. Therefore, the wavelength in meters is: $$\text{Wavelength} = 632.8\times10^{-9}~\mathrm{m}$$
02

Calculate the energy of a single photon

To calculate the energy of a single photon, we can use the Planck's constant \((h)\) and the speed of light \((c)\), with the following formula: $$E_{photon} = \frac{hc}{\lambda}$$ Where \(E_{photon}\) is the energy of a single photon and \(\lambda\) is the wavelength. The speed of light \((c)\) is approximately \(3\times10^{8}~\mathrm{m/s}\), and the Planck's constant \((h)\) is approximately \(6.626\times 10^{-34}~\mathrm{Js}\). Plug in the values to get the energy of a single photon: $$ E_{photon} = \frac{(6.626\times 10^{-34}~\mathrm{Js})(3\times10^{8}~\mathrm{m/s})}{632.8\times10^{-9}~\mathrm{m}}$$
03

Calculate the total energy of the pulse

The total energy of the pulse can be calculated using the power of the laser and the duration of the pulse: $$E_{total} = \text{Power} \times \text{Duration}$$ Given the power of the laser is \(0.5~\mathrm{W}\) and the duration of the laser pulse is \(2 \times 10^{-2}~\mathrm{s}\). Plug in these values to get the total energy of the pulse: $$E_{total} = (0.5~\mathrm{W}) (2\times10^{-2}~\mathrm{s})$$
04

Calculate the number of photons

To find the number of photons, divide the total energy of the pulse by the energy of a single photon: $$N = \frac{E_{total}}{E_{photon}}$$ Plugging in the values derived in steps 2 and 3: $$N= \frac{(0.5~\mathrm{W})(2\times10^{-2}~\mathrm{s})}{(\frac{(6.626\times 10^{-34}~\mathrm{Js})(3\times10^{8}~\mathrm{m/s})}{632.8\times10^{-9}~\mathrm{m}})}$$ This corresponds to: (A) \(10^{10}\)\ (B) \(10^{15}\)\ (C) \(10^{20}\)\ (D) \(10^{25}\)\ (E) \(10^{30}\) Calculating the value of \(N\) using the above equation, we get \(N \approx 10^{15}\). Therefore, the answer is (B) \(10^{15}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Laser Physics
Laser physics is the branch of optics that deals with the understanding and application of lasers. Lasers, which stand for Light Amplification by Stimulated Emission of Radiation, emit light through a process of optical amplification based on the stimulated emission of electromagnetic radiation. Unlike ordinary light sources, which emit a broad spectrum of light, lasers emit light in a specific, coherent wavelength, making them highly focused and precise.

This coherent light is particularly useful in various applications such as retinal surgery, where lasers must be precise enough to target very small areas without damaging surrounding tissue. In our exercise, the laser emits photons at a precise wavelength of 632.8 nm. The laser's precision and the ability to generate high intensity in a directed beam are quintessentially what makes laser physics an essential field in modern medicine and technology.
Photon Energy Calculation
The energy of a photon is calculated based on its wavelength or frequency. This energy determination is critical in fields such as quantum mechanics, astronomy, and any science involving electromagnetic radiation. For example, in the exercise, to determine the energy of photons emitted by a laser for retinal surgery, we use the formula:

\(E_{photon} = \frac{hc}{\lambda}\)

where \(E_{photon}\) is the energy of a single photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the photon's wavelength. This formula underpins the quantum mechanical relationship between energy and electromagnetic waves, illustrating that light has both wave-like and particle-like properties, a central concept in quantum physics.
Planck's Constant Usage
Planck's constant (\(h\)) is a fundamental constant in physics that plays a crucial role in quantum mechanics. It is the proportionality constant that relates the energy (\(E\)) of a photon to its frequency (\(u\)): \(E = hu\). It's also involved in the equation for calculating photon energy when the wavelength is known, as seen in the practice problem.

Planck's constant has a value of approximately \(6.626 \times 10^{-34}~\mathrm{Js}\), which indicates the very small energy scales typical in quantum processes. It forms the bedrock for much of our understanding of atomic and subatomic processes, including the behavior of photons in laser light. Moreover, Planck's constant usage not only includes energy calculations but also operations such as quantifying action in physical systems, establishing a limit to the precision with which certain pairs of physical properties, known as complementary variables, like position and momentum, can be known simultaneously – a concept known as the Heisenberg uncertainty principle.

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Most popular questions from this chapter

A photon of energy \(1.75 \mathrm{eV}\) has a wavelength of most nearly (A) \(710 \mathrm{~nm}\) (B) \(700 \mathrm{~nm}\) (C) \(650 \mathrm{~nm}\) (D) \(600 \mathrm{~nm}\) (E) \(550 \mathrm{~nm}\)

In 1923 Arthur Compton observed that \(x\)-rays scattered off free electrons were shifted in wavelength by an amount that could be explained by assuming that the \(\mathrm{x}\)-rays and electrons obeyed the relativistic relationships for energy and momentum. Compton used \(\mathrm{x}\)-rays \((\gamma)\) with a wavelength of approximately \(7.30 \times 10^{-11} \mathrm{~m}\). Assume that such a photon is incident on a stationary electron, as shown below. For purposes of illustration, the photon is reflected off the electron and its wavelength is observed to shift by a magnitude \(|\Delta \lambda|=\left|\lambda_o-\lambda_f\right|=h / m_e c\), where \(\lambda_o\) and \(\lambda_f\) are the initial and final wavelengths, respectively. a) What is the energy of the incoming photon in \(\mathrm{eV}\) ? in joules? b) Does the wavelength of the photon increase or decrease? Explain your reasoning. c) What is the momentum acquired by the electron?

Which of the following statements is true? The existence of the de Broglie wavelength \(\lambda_{d B}\) implies (A) that matter particles should undergo interference. (B) that matter waves travel at the speed of light. (C) that the frequency of matter waves is \(c / \lambda_{d B}\), where \(c\) is the speed of the particle. (D) that matter waves are given off by accelerating charges. (E) that matter waves are polarized.

An electron in the Bohr atom has an energy level determined by the radius of the orbit around the nucleus. The lowest energy state is given by the Bohr radius, which is roughly \(10^{-10} \mathrm{~m}\). The radius of a hydrogen nucleus is about \(10^{-14} \mathrm{~m}\). If the nucleus were the size of a basketball, the size of the atom would be nearest to (A) the size of a small town. (B) the size of a supermarket. (C) the size of a basketball court. (D) the size of a continent. (E) the size of the earth.

A hypothetical atom has energy levels at \(-12 \mathrm{eV},-8 \mathrm{eV},-3 \mathrm{eV},-1 \mathrm{eV}\). a) Draw the energy levels of the atom. Label the levels with the principal quantum number. b) An electron with velocity \(v=1.326 \times 10^6 \mathrm{~m} / \mathrm{s}\) is incident on the atom. What is the de Broglie wavelength of the electron? Ignore any relativistic effects. c) Can the electron excite an electron in any of the energy levels to a higher state? If so, which are the two levels involved? d) An electron decays from the \(-3 \mathrm{eV}\) state to the \(-8 \mathrm{eV}\) state and emits a photon. What is its wavelength? What part of the electromagnetic spectrum is it in?

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