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Chapter 28: Problem 44

A certain atom has an energy level \(3.50 \mathrm{eV}\) above the ground state. When excited to this state, it remains there for \(4.0 \mu \mathrm{s},\) on the average, before emitting a photon and returning to the ground state. (a) What is the energy of the photon? What is its wavelength? (b) What is the smallest possible uncertainty in energy of the photon?

Short Answer

Expert verified
(a) Energy: 3.50 eV, Wavelength: 354 nm. (b) Uncertainty: \( 1.32 \times 10^{-8} \mathrm{eV} \).

Step by step solution

01

Understanding the Energy of the Photon

The energy of the photon emitted when the atom transitions from the excited state back to the ground state is equal to the difference in the energy levels of the atom. Since the excited state is 3.50 eV above the ground state, the energy of the photon is also 3.50 eV.
02

Converting Photon Energy to Wavelength

To find the wavelength of the photon, we use the equation \( E = \frac{hc}{\lambda} \), where \( E \) is the energy of the photon in joules, \( h \) is Planck's constant \((6.626 \times 10^{-34} \mathrm{Js})\), \( c \) is the speed of light \((3.00 \times 10^{8} \mathrm{m/s})\), and \( \lambda \) is the wavelength. First, convert the energy from eV to joules using \( 1\mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J} \) giving \( E = 3.50 \times 1.602 \times 10^{-19} \mathrm{J} \). Then solve for \( \lambda \):\[\lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \mathrm{Js} \times 3.00 \times 10^{8} \mathrm{m/s}}{3.50 \times 1.602 \times 10^{-19} \mathrm{J}}\]Calculate \( \lambda \).
03

Calculate the Wavelength

After calculating, the resulting wavelength \( \lambda \) is approximately 354 nm, falling within the visible spectrum.
04

Uncertainty Principle for Energy

The uncertainty in the energy of the photon can be determined using the energy-time uncertainty principle: \( \Delta E \cdot \Delta t \geq \frac{h}{4\pi} \). Here, \( \Delta t = 4.0 \times 10^{-6} \mathrm{s} \). Solving for \( \Delta E \), we have:\[\Delta E \geq \frac{6.626 \times 10^{-34} \mathrm{Js}}{4\pi \times 4.0 \times 10^{-6} \mathrm{s}}\]Calculate \( \Delta E \) to find the minimum possible uncertainty.
05

Calculate the Uncertainty in Energy

Upon computation, the uncertainty \( \Delta E \) is approximately \( 1.32 \times 10^{-8} \mathrm{eV} \).

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Key Concepts

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

Energy Levels
In atomic physics, an energy level refers to the fixed energies that an electron within an atom can have. When an electron transitions between these levels, it absorbs or emits energy.
This absorption or emission results in the photon, the smallest energy carrier in this context.
  • The lowest energy level is called the ground state, while higher levels are excited states.
  • The difference in energy between these states determines the energy of any emitted or absorbed photon.
In the example, the energy level is 3.50 electron volts (eV) above the ground state, meaning the energy involved in the transition is 3.50 eV.
Photon Emission
Photon emission is a process in which an excited electron falls back to a lower energy level, emitting energy in the form of a photon. This is how atoms often return to their ground state.
In the given scenario, an electron in an excited state emits a photon with the same energy difference of 3.50 eV, as it transitions back to ground state.
  • Photons are packets of light energy.
  • The energy of a photon corresponds to the energy difference between the initial and final states of the electron.
This principle helps us understand many phenomena in atomic physics, from light emission in stars to lasers in technology.
Wavelength Calculation
Wavelength of light, or any electromagnetic wave, is inversely proportional to its energy. To calculate the wavelength of a photon, we apply the formula:
\[\lambda = \frac{hc}{E} \tag{1}\]where:
  • \( \lambda \) is the wavelength,
  • \( h \) is Planck鈥檚 constant \((6.626 \times 10^{-34} \mathrm{Js})\),
  • \( c \) is the speed of light \((3.00 \times 10^{8} \mathrm{m/s})\), and
  • \( E \) is the energy of the photon in joules.
First, we convert the energy from electron volts to joules by using the conversion factor
\(1 \, \mathrm{eV} = 1.602 \times 10^{-19} \, \mathrm{J}\).
Using this conversion, the energy is calculated, and formula (1) is applied to find the wavelength, yielding approximately 354 nm, visible light.
Uncertainty Principle
The uncertainty principle, formulated by Heisenberg, introduces the concept of intrinsic limits in measuring two complementary properties, like energy and time. It is expressed as:
\[\Delta E \cdot \Delta t \geq \frac{h}{4\pi} \tag{2}\]where:
  • \( \Delta E \) is the uncertainty in energy,
  • \( \Delta t \) is the uncertainty in time, and
  • \( h \) is Planck's constant.
In the problem, the average time the atom remains in the excited state is given as 4 microseconds. Using formula (2), we can determine the smallest possible uncertainty in the photon's energy.
This calculation shows that even with precise measurements, a fundamental uncertainty always remains, encapsulating a core principle in quantum physics.

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Most popular questions from this chapter

PRK surgery. Photorefractive keratectomy (PRK) is a laserbased surgery process that corrects near- and farsightedness by removing part of the lens of the eye to change its curvature and hence focal length. This procedure can remove layers \(0.25 \mu \mathrm{m}\) thick in pulses lasting \(12.0 \mathrm{~ns}\) with a laser beam of wavelength \(193 \mathrm{nm} .\) Low-intensity beams can be used because each individual photon has enough energy to break the covalent bonds of the tissue. (a) In what part of the electromagnetic spectrum does this light lie? (b) What is the energy of a single photon? (c) If a \(1.50 \mathrm{~mW}\) beam is used, how many photons are delivered to the lens in each pulse?

A pesky \(1.5 \mathrm{mg}\) mosquito is annoying you as you attempt to study physics in your room, which is \(5.0 \mathrm{~m}\) wide and \(2.5 \mathrm{~m}\) high. You decide to swat the bothersome insect as it flies toward you, but you need to estimate its speed to make a successful hit. (a) What is the maximum uncertainty in the horizontal position of the mosquito? (b) What limit does the Heisenberg uncertainty principle place on your ability to know the horizontal velocity of this mosquito? Is this limitation a serious impediment to your attempt to swat it?

A \(2.50 \mathrm{~W}\) beam of light of wavelength \(124 \mathrm{nm}\) falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is \(4.16 \mathrm{eV}\). Assume that each photon in the beam ejects an electron. (a) What is the work function (in electronvolts) of this metal? (b) How many photoelectrons are ejected each second from this metal? (c) If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part (b)? (d) If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part (b)?

An X-ray with a wavelength of \(0.100 \mathrm{nm}\) collides with an electron that is initially at rest. The X-ray's final wavelength is \(0.110 \mathrm{nm}\). What is the final kinetic energy of the electron?

An incident X-ray photon is scattered from a free electron that is initially at rest. The photon is scattered straight back at an angle of \(180^{\circ}\) from its initial direction. The wavelength of the scattered photon is \(0.0830 \mathrm{nm}\). (a) What is the wavelength of the incident photon? (b) What is the magnitude of the momentum of the electron after the collision? (c) What is the kinetic energy of the electron after the collision?
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