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 de Broglie wavelength of a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s?
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me = 9.11 × 10−31 kg).
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)
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27°C) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.
A electron of mass me revolves around a nucleus of charge +Ze. Show that it behaves like a tiny magnetic dipole. Hence prove that the magnetic moment associated wit it is expressed as `vecμ =−e/(2 m_e)vecL `, where `vec L` is the orbital angular momentum of the electron. Give the significance of negative sign.
State any one phenomenon in which moving particles exhibit wave nature.
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?
Sodium and copper have work function 2.3 eV and 4.5 eV respectively. Then, the ratio of the wavelengths is nearest to ______.
A proton and α-particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength λp to that λα is _______.
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?
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 proton, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as ______.
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 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.
For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?
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):
