Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
In this problem since both proton and α-particle are accelerated through the same potential difference,
We know that, `λ = h/sqrt(2mqv)`
∴ `λ oo 1/sqrt(mq)`
`λ_p/λ_α = sqrt(m_αq_α)/(m_pq_P) = sqrt(4m_p xx 2e)/sqrt(m_p xx e) = sqrt(8)`
∴ `λ_p = sqrt(8)λ_α`
i.e., wavelength of proton is `sqrt(8)` times wavelength of α-particle.
Important point: De-Broglie wavelength associated with the charged particles: The energy of a charged particle accelerated through potential difference V is `E = 1/2 mv^2 = qV`
Hence de-Broglie wavelength `λ = h/p = h/sqrt(2mE) = h/sqrt(2mqV)`
`λ_("Electron") = 12.27/sqrt(V) Å, λ_("Proton") = 0.286/sqrt(V) Å`
`λ_("Deutron") = 0.202/sqrt(V) Å, λ_(a"-particle") = 0.101/sqrt(V) Å`
APPEARS IN
संबंधित प्रश्न
A proton and an α-particle have the same de-Broglie wavelength Determine the ratio of their speeds.
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 ball of mass 0.060 kg moving at a speed of 1.0 m/s?
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.
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.
Describe briefly how the Davisson-Germer experiment demonstrated the wave nature of electrons.
Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
What are matter waves?
A proton and α-particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength λp to that λα is _______.
A proton, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as ______.
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 (mass m) with an initial velocity `v = v_0hati` is in an electric field `E = E_0hatj`. If λ0 = h/mv0, it’s de Broglie wavelength at time t is given by ______.
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 move at a right angle to each other. Their de-Broglie wavelengths are λ1 and λ2 respectively. The particles suffer a perfectly inelastic collision. The de-Broglie wavelength λ, of the final particle, is given by ______.
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:
Matter waves are ______.
