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
Describe the construction of photoelectric cell.
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 bullet of mass 0.040 kg travelling at the speed of 1.0 km/s?
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).
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)
State any one phenomenon in which moving particles exhibit wave nature.
Sodium and copper have work function 2.3 eV and 4.5 eV respectively. Then, the ratio of the wavelengths is nearest to ______.
The wavelength of the matter wave is dependent on ______.
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 ______
A particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is proportional to ______.
An electron is moving with an initial velocity `v = v_0hati` and is in a magnetic field `B = B_0hatj`. Then it’s de Broglie wavelength ______.
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.
Which of the following graphs correctly represents the variation of a particle momentum with its associated de-Broglie wavelength?
Matter waves are ______.
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):
