Advertisements
Advertisements
प्रश्न
Two particles A1 sand A2 of masses m1, m2 (m1 > m2) have the same de Broglie wavelength. Then ______.
- their momenta are the same.
- their energies are the same.
- energy of A1 is less than the energy of A2.
- energy of A1 is more than the energy of A2.
पर्याय
a and c
a and d
c and d
a and b
Advertisements
उत्तर
a and c
Explanation:
We know that de-Broglie wavelength `λ = h/(mv)`
Where, mv = p (momentum) of the particle
⇒ But we can express wavelength `λ = h/p` ⇒ `p = h/λ`
Hence, `p oo 1/λ ⇒ p_1/p_2 = λ_2/λ_1`
But particles have the same de-Broglie wavelength `(λ_1 = λ_2) = λ`
Then `p_1/p_2 = λ/λ` = 1 ⇒ `p_1 = p_2`
Thus, their momenta are the same.
Also, `E = 1/2 mv^2 = 1/2 mv^2 m/m`
= `1/2 (m^2v^2)/m = 1/2 p^2/m`
As p is constant, `E oo 1/m`
∴ `E_1/E_2 = m_2/m_1 < 1 ⇒ E_1 < E_2`
Important points:
Some important characteristics of Matter Waves:
- Matter wave represents the probability of finding a particle in space.
- Matter waves are not electromagnetic in nature.
- de-Broglie or matter wave is independent of the charge on the material particle. It means matter-wave of de-Broglie wave is associated with every moving particle (whether charged or uncharged).
- Practical observation of matter waves is possible only when the de-Broglie wavelength is of the order of the size of the particles.
- Electron microscope works on the basis of de-Broglie waves.
- The phase velocity of the matter waves can be greater than the speed of light.
- Matter waves can propagate in a vacuum, hence they are not mechanical waves.
- The number of de-Broglie waves associated with nth orbital electron is n.
- Only those circular orbits around the nucleus are stable whose circumference is an integral multiple of de-Broglie wavelength associated with the orbital electron.
APPEARS IN
संबंधित प्रश्न
Calculate the momentum of the electrons accelerated through a potential difference of 56 V.
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron, would have the same de Broglie wavelength.
Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn= 1.675 × 10−27 kg)
Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27°C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
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.
What are matter waves?
An electron (mass m) with an initial velocity `v = v_0hati (v_0 > 0)` is in an electric field `E = - E_0hati `(E0 = constant > 0). It’s de Broglie wavelength at time t is given by ______.
Relativistic corrections become necessary when the expression for the kinetic energy `1/2 mv^2`, becomes comparable with mc2, where m is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?
- λ = 10 nm
- λ = 10–1 nm
- λ = 10–4 nm
- λ = 10–6 nm
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 is accelerated from rest through a potential difference of 100 V. Find:
- the wavelength associated with
- the momentum and
- the velocity required by the electron.
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 ______.
Which of the following graphs correctly represents the variation of a particle momentum with its associated de-Broglie wavelength?
How will the de-Broglie wavelength associated with an electron be affected when the accelerating potential is increased? Justify your answer.
E, c and `v` represent the energy, velocity and frequency of a photon. Which of the following represents its wavelength?
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):
