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प्रश्न
Which of the following graphs correctly represents the variation of a particle momentum with its associated de-Broglie wavelength?
पर्याय
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उत्तर
Explanation:
de-Broglie wavelength is the wavelength that is connected to an object in relation to its momentum and mass. Typically, the force of a particle is inversely proportional to its de-Broglie wavelength.
A photon's momentum is determined by:
P = `E/c = h/lambda` where E is the energy of the photon, c is the speed of light in vacuum, h is Planck's constant and λ is the de-Broglie wavelength.
According to de-Broglie, `p = h/lambda` or `p ∝ 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 λ shall be a rectangular hyperbola.
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The graph which shows the variation of `(1/lambda^2)` and its kinetic energy, E is (where λ is de Broglie wavelength of a free particle):
