Question

Which of the following is true for photons?
A. \[E = \dfrac{{hc}}{\lambda }\]
B. \[E = \dfrac{1}{2}m{u^2}\]
C. \[P = \dfrac{E}{{2v}}\]
D. \[E = \dfrac{1}{2}m{c^2}\]

Answer 1

Hint: Photon is the smallest packet (quanta) of energy. By the particle nature of light, we can say that the light behaves as particles and this is confirmed by photoelectric effect, but according to the wave nature of light, light behaves as waves and is confirmed by phenomena like reflection, refraction etc. Now using this Planck quantum formula, we can solve this problem.

Formula Used:
To find the energy of the photon we have,
\[E = h\nu \]
Where, h is Planck’s constant and \[\nu \] is frequency of photon.

Complete step by step solution:
The energy of the photon is defined as the energy carried out by a single photon. The amount of energy of this photon is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. The higher the photon's frequency, the higher will be its energy.

We know that the energy of the photon is given by,
\[E = h\nu \]…… (1)
Since, frequency of photon can be written as,
\[\nu = \dfrac{c}{\lambda }\]
Here, c is the speed of light and \[\lambda \] is the wavelength of the photon.
Therefore, equation (1) can be written as,
\[E = \dfrac{{hc}}{\lambda }\]
Therefore, the energy of the photon is, \[E = \dfrac{{hc}}{\lambda }\].

Hence, Option A is the correct answer.

Note:Remember that, while solving the problems if the energy is given in eV, then we first need to convert it into Joules and then have to use this formula. But suppose if we want to find the wavelength and the frequency both then we could use the formula directly.