Advertisements
Advertisements
Question
Which of the following graphs correctly represents the variation of a particle momentum with its associated de-Broglie wavelength?
Options
Advertisements
Solution
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.
RELATED QUESTIONS
What is the de Broglie wavelength of a dust particle of mass 1.0 × 10−9 kg drifting with a speed of 2.2 m/s?
When a light wave travels from air to glass ______.
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?
An electromagnetic wave of wavelength ‘λ’ is incident on a photosensitive surface of negligible work function. If ‘m’ mass is of photoelectron emitted from the surface has de-Broglie wavelength λd, then ______
The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is `v_e = c/100`. Then ______.
- `E_e/E_p = 10^-4`
- `E_e/E_p = 10^-2`
- `p_e/(m_ec) = 10^-2`
- `p_e/(m_ec) = 10^-4`
A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values λ1, λ2 with λ1 > λ2. Which of the following statement are true?
- The particle could be moving in a circular orbit with origin as centre.
- The particle could be moving in an elliptic orbit with origin as its focus.
- When the de Broglie wavelength is λ1, the particle is nearer the origin than when its value is λ2.
- When the de Broglie wavelength is λ2, the particle is nearer the origin than when its value is λ1.
A particle A with a mass m A is moving with a velocity v and hits a particle B (mass mB) at rest (one dimensional motion). Find the change in the de Broglie wavelength of the particle A. Treat the collision as elastic.
An alpha particle is accelerated through a potential difference of 100 V. Calculate:
- The speed acquired by the alpha particle, and
- The de-Broglie wavelength is associated with it.
(Take mass of alpha particle = 6.4 × 10−27 kg)
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:
