Question

The energy of photon of wavelength λ is:
A.) \[\dfrac{c\lambda }{h}\]
B.) \[\dfrac{h\lambda }{c}\]
C.) \[\dfrac{hc}{\lambda }\]
D.) \[\dfrac{\lambda }{hc}\]

Answer 1

Hint: Photon is a discrete particle of light and travels with the velocity of light. Use this concept and apply it in the formula for energy-frequency relationship, also known as Plank’s-Einstein relation. Also use the frequency-wavelength relationship. As the options are given in terms of Planck's constant, wavelength and speed of light.

Formula used: Formula for Planck’s-Einstein relation
\[\text{E=h}\nu \]
Formula for wavelength-frequency relationship:
\[\nu \lambda =c\]

Complete step by step answer:
Photon is a discrete particle and has no mass. Light transports energy and this energy is proportional to the square of amplitude of the electric field of the light wave. This energy when it reaches the receiving end is not in continuous form but in discrete form and hence photons are discrete in nature. Thus, photons are the light particles.

There are few characteristics of photons:

A. They travel with the speed of light.
B. They are electrically neutral.
C. They are massless.
D. They can be created and even be destroyed.

Planck-Einstein relation determines the relation between the energy of each packet also known as quanta of light, the frequency of light and Planck's constant. The following relation holds:
\[\text{E=h}\nu \]… (1)
Where, E is the energy of each quanta, measured in Joules; \[\nu \] is the frequency, measured in Hertz; c is the velocity of light; h is called as Planck’s constant whose value is \[h=\text{ }6.626\text{ }\times \text{ }{{10}^{-34}}\text{ }Js\] .
If λ is the wavelength of light wave, we know the relationship of wavelength with frequency:
\[\nu \lambda =c\] .
Therefore equation (1) becomes:
\[\text{E=}\dfrac{\text{hc}}{\lambda }\]
From this photon energy equation, we can infer that the energy of a photon is always inversely related to the wavelength of the wave. Hence, the shorter the wavelength, the more the energy of the photon particle and if the wavelength of wave is longer, then the photon energy is higher.
Therefore, the correct answer is option B.

Note: This question can be analysed by dimensional analysis as well. That is if the dimensions, or even units of the quantities are known, the option with the dimension matching the quantity in the question is correct. For example, the unit of h is Js, $\lambda $ is m and c is m/s. Only option C. gives the units as J (Joules).