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
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
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)
70 cal of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from 30°C to 35°C. The amount of heat required to raise the temperature of the gas through the same range at constant volume will be (assume R = 2 cal/mol-K).
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 particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is proportional to ______.
A proton and an α-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λp and λa related to each other?
Assuming an electron is confined to a 1 nm wide region, find the uncertainty in momentum using Heisenberg Uncertainty principle (∆x∆p ≃ h). You can assume the uncertainty in position ∆x as 1 nm. Assuming p ≃ ∆p, find the energy of the electron in electron volts.
Two particles A and B of de Broglie wavelengths λ1 and λ2 combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional).
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)
A particle of mass 4M at rest disintegrates into two particles of mass M and 3M respectively having non zero velocities. The ratio of de-Broglie wavelength of particle of mass M to that of mass 3M will be:
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 ______.
