Advertisements
Advertisements
प्रश्न
Relativistic corrections become necessary when the expression for the kinetic energy `1/2 mv^2`, becomes comparable with mc2, where m is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?
- λ = 10 nm
- λ = 10–1 nm
- λ = 10–4 nm
- λ = 10–6 nm
पर्याय
a and c
a and d
c and d
a and b
Advertisements
उत्तर
c and d
Explanation:
De-Broglie or matter wave is independent of die charge on the material particle. It means matter wave of the de-Broglie wave is associated with every moving particle (whether charged or uncharged).
The de-Broglie wavelength at which relativistic corrections become important is that the phase velocity of the matter waves can be greater than the speed of the light (3 × 108 m/s).
The wavelength of de-Broglie wave is given by λ = h/p = h/mv
Here, h = 6.6 × 10-34 Js
And for electron, m = 9 × 10-31 kg
To approach these types of problems we use the hit and trial method by picking up each option one by one.
In option (a), λ1 = 10 nm = 10 × 10–9 m = 10–8 m
⇒ `v_1 = (6.6 xx 10^-34)/((9 xx 10^-31) xx 10^-8)`
= `2.2/3 xx 10^5 = 10^5` m/s
In option (b), λ2 = 10–1 nm = 10–1 × 10–9 m = 10–10 m
⇒ `v_2 = (6.6 xx 10^-34)/((9 xx 10^-31) xx 10^-10) = 10^7` m/s
In option (c), λ3 = 10–4 nm = 10–4 × 10–9 m = 10–13 m
⇒ `v_3 = (6.6 xx 10^-34)/((9 xx 10^-31) xx 10^-13) = 10^10` m/s
In option (d), λ4 = 10–6 nm = 10–6 × 10–9 m = 10–15 m
⇒ `v_4 = (6.6 xx 10^-34)/((9 xx 10^-31) xx 10^-15) = 10^12` m/s
Thus options (c) and (d) are correct as v3 and v4 is greater than 3 × 108 m/s.
APPEARS IN
संबंधित प्रश्न
Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
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?
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.
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)
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.
Compute the typical de Broglie wavelength of an electron in a metal at 27°C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10−10 m.
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?
State any one phenomenon in which moving particles exhibit wave nature.
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.
Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
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 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 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?
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).
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 ______.
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 ______.
How will the de-Broglie wavelength associated with an electron be affected when the accelerating potential is increased? Justify your answer.
Matter waves are ______.
