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**Hint:**The relation between de Broglie wavelength and linear momentum is calculated from the energy of the photon. De Broglie’s equation equates de Broglie wavelength to the ratio of Planck’s constant and linear momentum.

**Formula Used:**The formulae used in the solution are given here.

Momentum of a photon is given by-

$P = \dfrac{E}{c} = \dfrac{h}{\lambda }$ where $E$ is the energy of the photon, $c$ is the speed of light in vacuum, $h$ is the Planck’s constant and $\lambda $ is the de Broglie wavelength.

**Complete Step by Step Solution:**The wavelength that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength. A particle’s de Broglie wavelength is usually inversely proportional to its force.

Momentum of a photon is given by-

$P = \dfrac{E}{c} = \dfrac{h}{\lambda }$ where $E$ is the energy of the photon, $c$ is the speed of light in vacuum, $h$ is the Planck’s constant and $\lambda $ is the de Broglie wavelength.

According to de Broglie, $p = \dfrac{h}{\lambda }$ or $p\alpha \dfrac{1}{\lambda }$.

By this relation we can conclude that the linear momentum of a photon is inversely proportional to the de Broglie wavelength. The graph of $p$ vs $\lambda$ shall be a rectangular hyperbola.

It will look like,

**Hence, the correct answer is Option D.**

**Note:**It is said that matter has a dual nature of wave-particles. De Broglie waves named after the discoverer Louis de Broglie, is the property of a material object that varies in time or space while behaving similar to waves. It is also called matter-waves. It holds great similarity to the dual nature of light which behaves as particle and wave, which has been proven experimentally.

The physicist Louis de Broglie suggested that particles might have both wave properties and particle properties. The wave nature of electrons was also detected experimentally to substantiate the suggestion of Louis de Broglie.

The objects which we see in day-to-day life have wavelengths which are very small and invisible, hence, we do not experience them as waves. However, de Broglie wavelengths are quite visible in the case of subatomic particles.

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