Question

Can you calculate the energy of a photon of wavelength 11.56 meters, (Planck’s constant is $6.626\text{ }\times \text{ }{{10}^{-34}}$ joule seconds; the speed of light is $2.998\text{ }\times \text{ }{{10}^{8}}$ m/s)?

Answer 1

A sort of elementary particle is the photon. It is the force carrier for the electromagnetic force and the quantum of the electromagnetic field, which includes electromagnetic radiation such as light and radio waves. Because photons have no mass, they always travel at the speed of light in vacuum, which is 299792458 m/s (about 186,282 mi/s). The photon belongs to the boson family of particles.
We calculate the energy of photons here.

Complete answer:
The wavelength of a periodic wave is its spatial period, or the distance over which the wave's form repeats. It's the distance between two adjacent corresponding points of the same phase on the wave, such two adjacent crests, troughs, or zero crossings, and it's a feature of both travelling and standing waves, as well as other spatial wave patterns. The spatial frequency is the inverse of the wavelength. The Greek letter lambda () is often used to represent wavelength. Modulated waves, as well as the sinusoidal envelopes of modulated waves or waves generated by the interference of multiple sinusoids, are sometimes referred to as wavelengths.
The photon's electromagnetic frequency is directly proportional to the amount of energy it contains, and hence the wavelength is inversely proportional. The greater the frequency of a photon, the more energy it has. In other words, the longer the wavelength of a photon, the lower its energy.
We can use the formula $E=\dfrac{h c}{\lambda}$ , where $h$ is Planck's Constant, $c$ is the speed of the light, and $\lambda$ is wavelength of the photon.
$E=\dfrac{h c}{\lambda}=\dfrac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s} \times 2.998 \times 10^{8} \mathrm{~m} \cdot \mathrm{s}^{-1}}{11.56 \mathrm{~m}}=1.718 \times 10^{-26} \mathrm{~J}$
Hence The energy of the photon is $1.718 \times 10^{-26} \mathrm{~J}$ .

Note:
Any unit of energy can be used to express photon energy. The electronvolt (eV) and the joule are two popular units for measuring photon energy (as well as its multiples, such as the microjoule). Because one joule = $6.24\times {{10}^{18}}\text{ }eV$ , the bigger units may be more useful in describing the energy of photons with greater frequency and energy, such as gamma rays, as opposed to photons with lower energy, such as those in the radio frequency part of the electromagnetic spectrum.