Advertisements
Advertisements
प्रश्न
Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Advertisements
उत्तर
Potential difference, V = 56 V
Planck’s constant, h = 6.6 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
Charge on an electron, e = 1.6 × 10−19 C
De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:
`lambda = 12.27/sqrt("V")` Å
= `12.27/sqrt56 xx 10^(-10) "m"`
= 0.1639 nm
Therefore, the de Broglie wavelength of each electron is 0.1639 nm.
APPEARS IN
संबंधित प्रश्न
What is the de Broglie wavelength of a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s?
An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
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.
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).
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.
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?
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.
Sodium and copper have work function 2.3 eV and 4.5 eV respectively. Then, the ratio of the wavelengths is nearest to ______.
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, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as ______.
The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is `v_e = c/100`. Then ______.
- `E_e/E_p = 10^-4`
- `E_e/E_p = 10^-2`
- `p_e/(m_ec) = 10^-2`
- `p_e/(m_ec) = 10^-4`
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).
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.
The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1 : `sqrt2`. Then, the ratio of Vp to Vd will be ______.
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:
For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?
How will the de-Broglie wavelength associated with an electron be affected when the velocity of the electron decreases? Justify your answer.
How will the de-Broglie wavelength associated with an electron be affected when the accelerating potential is increased? Justify your answer.
