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Chapter 2: Problem 2

Show that the energy of a photon in electron-volts, \(E\), is related to its wavelength in nano-meters, \(\lambda\), by the equation \(E=1240 / \lambda\).

Short Answer

Expert verified
The energy of a photon in eV is given by \( E = \frac{1240}{\lambda} \), with \( \lambda \) in nm.

Step by step solution

01

Understand the Photon Energy Formula in Joules

The energy of a photon is given by the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{Js}) \), \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \), and \( \lambda \) is the wavelength in meters.
02

Convert Energy from Joules to Electron-Volts

1 electron-volt (eV) equals \( 1.602 \times 10^{-19} \, \text{J} \). To convert the energy from Joules to electron-volts, use the transformation: \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \).
03

Convert Wavelength to Meters

Given \( \lambda \) in nano-meters, first convert it to meters: \( 1 \, \text{nm} = 1 \times 10^{-9} \, \text{m} \). Thus, \( \lambda = \lambda \times 10^{-9} \, \text{m} \).
04

Substitute Constants and Simplify

Starting from \( E = \frac{hc}{\lambda} \) in Joules, substitute \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \) to obtain \( E(\text{J}) = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{\lambda \times 10^{-9}} \). Simplifying gives \( E(\text{J}) = \frac{1.988 \times 10^{-25}}{\lambda \times 10^{-9}} \).
05

Convert Energy to Electron-Volts

Divide the energy in Joules by the conversion factor: \( E(\text{eV}) = \frac{1.988 \times 10^{-25}}{\lambda \times 10^{-9}} \times \frac{1}{1.602 \times 10^{-19}} \). Simplifying gives \( E = \frac{1240}{\lambda} \) in eV for \( \lambda \) in nm.

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Key Concepts

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

Wavelength
Wavelength represents the distance between two consecutive peaks of a wave. This wave characteristic applies to light, sound, and other types of waves. When we talk about the wavelength of light, we're referring to the physical length of one cycle of the electromagnetic wave. This length is usually measured in meters, but for convenience, when discussing light, wavelength is often given in nanometers (nm) because the values are tiny.
A nanometer is a billionth of a meter (1 nm = 10^{-9} meters), making it a suitable unit for describing the dimensions of even the smallest particles like photons.
To convert the wavelength from nanometers to meters, multiply by 10^{-9}. This conversion is crucial when working with formulas that compute photon energy, helping ensure units are consistent and calculations accurate.
Electron-Volts
An electron-volt (eV) is a unit of energy specifically useful in the fields of atomic and particle physics. It represents the amount of kinetic energy gained or lost by an electron when it is accelerated through an electric potential difference of one volt.
It's a smaller unit well-suited for the tiny energy changes occurring at the atomic scale, where more commonly used units like joules would be inconveniently small.
Here's how it relates to joules:
  • 1 eV = 1.602 脳 10^{-19} joules
This means that a single electron-volt is a fraction of a joule, allowing scientists to more easily discuss and calculate the energy changes happening in particles and photons in physics. By converting energy values to electron-volts, we simplify discussions and calculations focused on atomic and subatomic processes, where these values matter most.
Planck's Constant
Planck鈥檚 constant is a fundamental feature in quantum mechanics. It offers the bridge between the wave and particle nature of light. The constant is denoted as \( h \) and valued at \( 6.626 \times 10^{-34} \, \text{Js} \). This small number is significant because it allows us to compute the discrete energy packets, or quanta, each photon carries.
In the calculation of photon energy, Planck's constant is a crucial part. According to the equation \( E = \frac{hc}{\lambda} \), Planck's constant (\( h \)) pairs with the speed of light (\( c \)) to relate energy to wavelength. This reveals that a photon's energy is inversely proportional to its wavelength: as the photon鈥檚 wavelength shortens, its energy increases.
Understanding Planck's constant helps us appreciate why light behaves as both a wave and a particle, a centerpiece of quantum theory and modern physics.

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Most popular questions from this chapter

The frequency of the beats generated by waves of frequency \(f\) and \(f\) is \(\left|f-f^{\prime}\right| .\) In order to be certain that \(f=f^{\prime}\) how long must we wait to see whether beats are generated or not? And after a period of time \(\Delta t\), with what accuracy could we declare that the two frequencies are equal?

a. Calculate the difference, \(\Delta E_{n}=E_{n+1}-E_{n}\), between the energy levels of a particle of mass \(m\) that is trapped in an infinite potential well of length \(L\) in the case where \(n \gg 1\) b. Compare the values of \(\Delta E_{n}\) and \(\Delta E_{n+1}\) in the case where \(n \gg 1\). Comment on the result. c. Find an expression for the ratio \(\Delta E_{n} / E_{n}\). How does the value of this ratio change as \(n\) grows larger?

The maximum kinetic energy of the photoelectrons emitted from a tungsten plate illuminated by light of wavelength \(248 \mathrm{~nm}\) is \(8.6 .10^{-20} \mathrm{~J}\). Calculate the work function of tungsten.

The threshold frequency for the emission of photoelectrons from copper is \(1.1 \cdot 10^{15} \mathrm{~Hz}\). Calculate the maximum energy - in Joules and in electron-volts - of the photoelectrons emitted from a copper plate by radiation of frequency \(15 \cdot 10^{15} \mathrm{~Hz}\).

The frequency, \(f\), of a harmonic oscillator of mass \(m\) and elasticity constant \(k\) is given by the equation \(f=\frac{1}{2 \pi} \sqrt{k / m}\). The energy of the oscillator is given by \(E=p^{2} / 2 m+k x^{2} / 2\), where \(p\) is the system's linear momentum and \(x\) is the displacement from its equilibrium position. Use the uncertainty principle, \(\Delta x \cdot \Delta p \approx \hbar / 2\), to express the oscillator's energy \(E\) in terms of \(x\) and show, by taking the derivative of this function and setting \(d E / d x=0\), that the minimum energy of the oscillator (its ground state energy) is \(E_{\min }=h f / 2\).
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