Question

Wavelength of photon having energy \(1 \mathrm{eV}\) would be (a) \(1.24 \times 10^{-4} \mathrm{~m}\) (b) \(1.24 \times 10^{-6} \mathrm{~m}\) (c) \(1.24 \times 10^{-5} \mathrm{~m}\) (d) \(1.24 \times 10^{4} \mathrm{~m}\)

Short answer

The wavelength of a photon having an energy of 1 eV is (b) \(1.24 \times 10^{-6} \text{ m}\).

Step-by-step solution

Use the energy-wavelength relationship

The relationship between the energy (E) of a photon and its wavelength (\( \lambda \)) is given by the equation \( E = \frac{hc}{\lambda} \), where 'h' is the Planck constant \( (6.626 \times 10^{-34} \text{ Js}) \) and 'c' is the speed of light in vacuum \( (3 \times 10^{8} \text{ m/s}) \). The energy given is 1 eV, which needs to be converted to Joules (J) using the conversion factor \( 1 \text{eV} = 1.602 \times 10^{-19} \text{J} \).

Convert the photon's energy to Joules

Using the conversion factor, the energy in Joules is \( E = 1 \text{eV} \times 1.602 \times 10^{-19} \text{J/eV} = 1.602 \times 10^{-19} \text{ J} \).

Rearrange the equation to solve for wavelength

Rearrange the energy-wavelength equation to solve for \( \lambda \). \( \lambda = \frac{hc}{E} \).

Plug in the values and calculate the wavelength

Now, plug in the values of h, c, and E into the equation to find the wavelength. \( \lambda = \frac{(6.626 \times 10^{-34} \text{ Js})(3 \times 10^{8} \text{ m/s})}{(1.602 \times 10^{-19} \text{ J})} = \frac{1.9878 \times 10^{-25} \text{ Jm/s}}{1.602 \times 10^{-19} \text{ J}} \). Performing the division gives \( \lambda = 1.24 \times 10^{-6} \text{ m} \).

Key concepts

Energy-Wavelength Relationship

The energy-wavelength relationship is a cornerstone concept in understanding the behavior of photons. Photons, which are the basic units of light, carry energy that is inversely proportional to their wavelength. The formula that ties these properties together is quite elegant and simple:
\[\begin{equation} E = \frac{hc}{\lambda} \end{equation}\]
In this equation, \( E \) represents the energy of the photon measured in Joules (J), \( \lambda \) is the wavelength measured in meters (m), \( h \) is the Planck constant (which we will discuss shortly), and \( c \) is the speed of light in a vacuum.

Understanding this equation helps us grasp the conceptual framework of quantum mechanics where light, classically considered a wave, also behaves as a particle—each with a discrete packet of energy, known as a photon. The shorter the wavelength, the higher the energy of the photon, and vice versa. Thus, visible light has more energy than microwaves but less than X-rays.

Planck Constant

The Planck constant (denoted as \(h\)) is a fundamental constant in physics, particularly in quantum mechanics. Its value is
\[\begin{equation} h = 6.626 \times 10^{-34} \text{Js} \end{equation}\]
which indicates the proportionality factor between the minimum increment of energy, termed a quantum, and the frequency of its associated electromagnetic wave. In simpler terms, the Planck constant relates a photon’s energy to its frequency and thus also to its wavelength through the speed of light.

The discovery of the Planck constant was a pivotal moment in physics, leading to the birth of quantum theory. It represents the smallest action in nature and sets a fundamental scale for the quantum behavior of particles. For example, in our context, it’s the linchpin allowing us to calculate the energy of a photon given its wavelength—or vice versa.

Speed of Light

The speed of light in a vacuum, commonly denoted as \(c\), is another foundational constant in physics. Its value is a staggering
\[\begin{equation} c = 3 \times 10^8 \text{m/s} \end{equation}\]
It's important to note that the speed of light is the same for all observers, regardless of the motion of the light source or observer. This constancy underpins many theories in physics and is central to the theory of relativity. In the context of photons and their energy, \(c\) allows us to connect the frequency and wavelength of electromagnetic radiation, since
\[\begin{equation} c = \lambda \times f \end{equation}\]
where \( f \) is the frequency. For educational purposes, remembering that light travels at this finite speed is essential for understanding phenomena such as the time delay of light from distant stars and the behavior of light in various media.

Energy Conversion eV to Joules

Energy can be expressed in various units, and two common ones are electronvolts (eV) and Joules (J). An electronvolt is the amount of kinetic energy gained by a single electron when it accelerates through an electric potential difference of one volt. In contrast, the Joule is the SI unit of energy and is more comprehensive, being used across all fields of physics.
For photon energy, the conversion factor between these units is essential:
\[\begin{equation} 1 \text{eV} = 1.602 \times 10^{-19} \text{J} \end{equation}\]
This conversion allows us to switch between the two units depending on the context. For practical purposes, such as in semiconductor physics or when dealing with particles like electrons, energy is usually discussed in electronvolts due to the more manageable numbers. However, when performing calculations for photons, as in our exercise, it is converted into Joules to apply the Planck-Einstein relation for energy and wavelength.