Advertisements
Advertisements
प्रश्न
A proton, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as ______.
पर्याय
λp = λn > λe > λα
λα < λp = λn > λe
λe < λp = λn > λα
λe = λp = λn = λα
Advertisements
उत्तर
A proton, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as `underline(λ_α < λ_p = λ_n > λ_e)`.
Explanation:
According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.
deiBroglie wavelength: According to de-Broglie theory, the wavelength of de-Broglie wave is given by
`λ = h/p = h/(mv) = h/sqrt(2mE) ⇒ λ oo 1/p oo 1/v oo 1/sqrt(E)`
Where h = Plank's constant, m = Mass of the particle, v = Speed of the particle, E = Energy of the particle.
The smallest wavelength whose measurement is possible is that of γ-rays.
The wavelength of matter waves associated with the microscopic particles like electron, proton neutron, α-particle etc is of the order of 10–10 m.
We conclude from above that the relation between de-Broglie wavelength λ and kinetic energy K of the particle is given by
`λ = h/sqrt(2mK)`
Here, for the given value of energy K, `h/sqrt(2K)` is a constant.
Thus, `λ oo 1/sqrt(m)`
∴ `λ_p : λ_n : λ_e : λ_α = 1/sqrt(m_p) : 1/sqrt(m_n) : 1/sqrt(m_e) : 1/sqrt(m_α)`
Comparing the wavelength of proton and neutron, mp = mn, hence λp = λn
As mα > mp therefore λα < λp
As me > mn therefore λe < λn
Hence. λα < λp = λn < λe
APPEARS IN
संबंधित प्रश्न
Describe the construction of photoelectric cell.
Calculate the momentum of the electrons accelerated through a potential difference of 56 V.
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
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?
For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10−10 m?
Find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of `(3/2)` kT at 300 K.
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
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.
Compute the typical de Broglie wavelength of an electron in a metal at 27°C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10−10 m.
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?
What are matter waves?
Sodium and copper have work function 2.3 eV and 4.5 eV respectively. Then, the ratio of the wavelengths is nearest to ______.
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 ______.
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 ______.
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.
Given below are two statements:
Statement - I: Two photons having equal linear momenta have equal wavelengths.
Statement - II: If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease.
In the light of the above statements, choose the correct answer from the options given below.
For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?
In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to ______.
How will the de-Broglie wavelength associated with an electron be affected when the velocity of the electron decreases? Justify your answer.
