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Chapter 42: Problem 2

Compute the energy of a photon of blue light of wavelength 450 \(\mathrm{nm} .\) $$ E=\frac{h \mathrm{c}}{\lambda}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{450 \times 10^{-9} \mathrm{~m}}=4.42 \times 10^{-19} \mathrm{~J}=2.76 \mathrm{eV} $$

Short Answer

Expert verified
The energy of a blue light photon with a wavelength of 450 nm is 4.42 脳 10鈦宦光伖 J or 2.76 eV.

Step by step solution

01

Understand the Formula

The formula to calculate the energy of a photon is given by \(E = \frac{hc}{\lambda}\), where \(E\) is the energy, \(h\) is Planck's constant \((6.63 \times 10^{-34} \text{ J} \cdot \text{s})\), \(c\) is the speed of light \((2.998 \times 10^{8} \text{ m/s})\), and \(\lambda\) is the wavelength. In this problem, we are given \(\lambda = 450 \text{ nm}\).
02

Convert Wavelength to Meters

Since the given wavelength is in nanometers, convert it to meters. 1 nanometer (nm) equals \(10^{-9}\) meters. Thus, \(450\, \text{nm} = 450 \times 10^{-9} \text{ m}\).
03

Plug Values into the Formula

Now substitute the values into the energy formula: \(E = \frac{(6.63 \times 10^{-34} \text{ J} \cdot \text{s})(2.998 \times 10^{8} \text{ m/s})}{450 \times 10^{-9} \text{ m}}\).
04

Calculate the Energy in Joules

Perform the calculations: \(E = \frac{(6.63 \times 10^{-34} \times 2.998 \times 10^{8})}{450 \times 10^{-9}}\), which simplifies to \(4.42 \times 10^{-19} \text{ J}\).
05

Convert Energy to Electronvolts

To convert energy from joules to electronvolts (eV), use the conversion factor \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\). Thus, \(E = \frac{4.42 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} = 2.76 \, \text{eV}\).

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Key Concepts

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

Planck's Constant
Planck's constant is a fundamental constant in physics denoted by the letter \( h \), with a value of \( 6.63 \times 10^{-34} \text{ J} \cdot \text{s} \). It plays a crucial role in the quantum mechanics field by bridging the gap between the energy of a photon and its associated frequency or wavelength. This constant serves as a key tool when calculating the energy of photons, such as when determining energy from the wavelength of light.
  • Definition: Planck's constant quantizes the energy levels of photons.
  • Importance: It confirms that energy is quantized, challenging classical mechanics.
This value ensures accuracy and consistency in photon energy calculations. The presence of Planck's constant in the equation \( E = \frac{hc}{\lambda} \) showcases its pivotal role in connecting light properties like wavelength to photon energy.
Wavelength Conversion
Understanding wavelength conversion is essential, especially when dealing with different units of measurement. Wavelength is often provided in nanometers (nm), but calculations typically require it in meters (m).
  • Conversion Factor: 1 nanometer equals \( 10^{-9} \) meters.
  • Example: To convert 450 nm to meters, multiply by \( 10^{-9} \): \[ 450 \text{ nm} = 450 \times 10^{-9} \text{ meters} = 4.5 \times 10^{-7} \text{ m} \]
Accurate conversion is crucial, as it directly affects the energy calculation if the wavelength is not correctly converted. This step ensures that subsequent calculations, particularly those involving physical constants that assume standard SI units, are mathematically sound.
Energy Calculation
The energy of a photon can be calculated using the relation \( E = \frac{hc}{\lambda} \). This formula requires the multiplication of Planck's constant \( h \) and the speed of light \( c \) and division by the wavelength \( \lambda \).
  • Formula: \[ E = \frac{(6.63 \times 10^{-34} \text{ J} \cdot \text{s})(2.998 \times 10^{8} \text{ m/s})}{450 \times 10^{-9} \text{ m}} \]
  • Calculation: This process results in an energy value of \( 4.42 \times 10^{-19} \text{ J} \).
This calculated energy provides insight into the photon鈥檚 strength or its ability to perform work or cause change, which is essential in applications like electronics, photonics, and quantum computing. This understanding allows you to explore more about how different wavelengths affect photon energy in various fields.
Electronvolt Conversion
An electronvolt (eV) is a unit of energy widely used in atomic and particle physics. It provides a convenient scale for quantifying the energy of particles, such as electrons or photons, in terms readily comprehensible in laboratory contexts.
  • Conversion Factor: 1 electronvolt is defined as \( 1.602 \times 10^{-19} \text{ J} \).
  • Conversion Process: Convert the photon energy from joules to electronvolts: \[ E = \frac{4.42 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} = 2.76 \text{ eV} \]
Using electronvolts streamlines calculations and comparisons in many physics contexts, making it easier to share and interpret results in studies and experiments involving particle energy.

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Most popular questions from this chapter

If you double the wavelength of a photon, what happens to its energy? Explain your answer.

Will photoelectrons be emitted by a metal surface, of work function \(4.4 \mathrm{eV}\), when illuminated by visible light? As in \(\underline{\text { Problem } 42.4}\), the released-electron's \(\mathrm{KE}=0\) and so $$ \text { eshold } \lambda=\frac{h \mathrm{c}}{\phi}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{4.4\left(1.602 \times 10^{-19}\right) \mathrm{J}}=282 \mathrm{~nm} $$ Hence, visible light \((350 \mathrm{~nm}\) to \(700 \mathrm{~nm})\) cannot eject photoelectrons from copper.

What potential difference must be applied to stop the fastest photoelectrons emitted by a nickel surface under the action of ultraviolet light of wavelength \(200 \mathrm{~nm}\) ? The work function of nickel is \(5.01 \mathrm{eV}\). $$ \mathrm{E}=\frac{h \mathrm{c}}{\lambda}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{2000 \times 10^{-10} \mathrm{~m}}=9.95 \times 10^{-19} \mathrm{~J}=6.21 \mathrm{eV} $$ Then, from the photoelectric equation, the energy of the fastest emitted electron is $$ 6.21 \mathrm{eV}-5.01 \mathrm{eV}=1.20 \mathrm{eV} $$ Hence, a negative retarding potential of \(1.20 \mathrm{~V}\) is required. This is the stopping potential.

In a process called pair production, a photon is transformed into an electron and a positron. A positron has the same mass \(\left(m_{e}\right)\) as the electron, but its charge is \(+e\). To three significant figures, what is the minimum energy a photon can have if this process is to occur? What is the corresponding wavelength? The electron-positron pair will come into existence moving with some minimum amount of KE. The particles will separate, and as they do they will slow down. When far apart each will have a mass of \(9.11 \times 10^{-31} \mathrm{~kg} .\) In effect, KE goes into \(\mathrm{PE}\), which is manifested as mass. Thus, the minimum energy photon at the start of the process must have the energy equivalent of the free-particle mass of the pair at the end of the process. Hence,

What are the speed and momentum of a 500 -nm photon?
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