Question

Which of the following formula represents the energy of a photon?
(a) $E = \dfrac{{hc}}{\lambda }$
(b) $E = hc\lambda$
(c) $E = \dfrac{{h\lambda }}{c}$
(d) $E = \dfrac{{c\lambda }}{h}$

Answer 1

Photons may be a discrete particle of light and travel with the velocity of light. Use this idea and apply it within the formula for an energy-frequency relationship, also called as Plank’s-Einstein relation. Also, use the frequency-wavelength relationship. Additionally, the physical quantities wavelength and frequency are inversely proportional to one another.

Formula used: The formula for Planck’s-Einstein relation
$E = hv$
The formula for wavelength-frequency relationship:
$v\lambda = c$

Complete step-by-step solution:
The photon may be a discrete particle and has no mass. Sunshine or Light transports energy and this energy is proportional to the square of the amplitude of the electrical field of the light wave. This energy when it reaches the receiving end isn’t in continuous form but discrete form and hence photons are discrete in nature. Thus, photons are light or sunshine particles.
There are few characteristics of photons- They travel with the speed of light, they’re electrically neutral, they’re massless, and that they will be created and even be destroyed.
Planck-Einstein's relation determines the relation between the energy of every packet also referred to as quanta of light, the frequency of light, and Planck's constant. The subsequent relation holds:
$E = hv$
Where, $E$ is the energy of each quantum, measured in Joules; $v$ is the frequency, measured in Hertz; $c$ is the velocity of light; $h$ is called as Planck’s constant whose value is $h = 6.626 \times {10^{ - 34}}Js$.
If $\lambda$ is the wavelength of lightwave, we know the relationship of wavelength with frequency:

v\lambda = c \\\ v = \dfrac{c}{\lambda } \\\ \end{gathered} $$ Therefore, $$E = \dfrac{{hc}}{\lambda }$$ From this photon energy equation, we are able to infer that the energy of a photon is often inversely associated with the wavelength of the wave. Hence, the shorter the wavelength, the more the energy of the photon particle, and if the wavelength of a wave is longer, then the photon energy is higher. **Therefore, the correct answer is option A.** **Note:** Photons are electrically neutral particles and are hence not deflected by magnetic and electric fields. Additionally, all photons of a specific frequency and wavelength will have identical energy and momentum independent of the intensity of radiation.