Advertisements
Advertisements
प्रश्न
Which of the following graphs correctly represents the variation of a particle momentum with its associated de-Broglie wavelength?
विकल्प
Advertisements
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.
संबंधित प्रश्न
What is the de Broglie wavelength of a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s?
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:
E = hv, p = `"h"/lambda`
But while the value of λ is physically significant, the value of v (and therefore, the value of the phase speed vλ) has no physical significance. Why?
The wavelength λ of a photon and the de-Broglie wavelength of an electron have the same value. Show that energy of a photon in (2λmc/h) times the kinetic energy of electron; where m, c and h have their usual meaning.
Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
What are matter waves?
The de-Broglie wavelength associated with a material particle when it is accelerated through a potential difference of 150 volt is 1 Å. What will be the de-broglie wavelength associated with the same particle when it is accelerated through a potential difference of 4500 V?
A proton, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as ______.
The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1 : `sqrt2`. Then, the ratio of Vp to Vd will be ______.
An electron of mass me, and a proton of mass mp = 1836 me are moving with the same speed. The ratio of the de Broglie wavelength `lambda_"electron"/lambda_"proton"` will be:
