English
Karnataka Board PUCPUC Science 2nd PUC Class 12

An electron is moving with an initial velocity v=v0i^ and is in a magnetic field B=B0j^. Then it’s de Broglie wavelength ______.

Advertisements
Advertisements

Question

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 ______.

Options

  • remains constant.

  • increases with time.

  • decreases with time.

  • increases and decreases periodically.

MCQ
Fill in the Blanks
Advertisements

Solution

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 remains constant.

Explanation:

If a particle is carrying a positive charge q and moving with a velocity v and enters a magnetic field 5 then it experiences a force F which is given by the expression

F = q(v × B) = $ F = qvB sin θ. As this force is perpendicular to v and B, so the magnitude of v will not change, i.e. momentum (p = mv) will remain constant in magnitude.

According to the problem, `vecv = v_0i` and `vecB = B_0j`

Magnetic force on moving electron = `-e[v_0i xx B_0j] ⇒ - ev_0B_0k`

As this force is perpendicular to `vecv` and `vecB`, so the magnitude of v will not change, i.e. momentum (p = mv) will remain constant in magnitude. Hence, de-Broglie wavelength `lambda = h/(mv)` remains constant.

shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Dual Nature Of Radiation And Matter - Exercises [Page 69]

APPEARS IN

NCERT Exemplar Physics Exemplar [English] Class 12
Chapter 11 Dual Nature Of Radiation And Matter
Exercises | Q 11.06 | Page 69

Video TutorialsVIEW ALL [2]

RELATED QUESTIONS

Calculate the momentum of the electrons accelerated through a potential difference of 56 V.


Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.


Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).


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.


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.


Which one of the following deflect in electric field


The wavelength of the matter wave is dependent on ______.


A proton and α-particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength λp to that λα is _______.


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 ______.


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 ______.


An alpha particle is accelerated through a potential difference of 100 V. Calculate:

  1. The speed acquired by the alpha particle, and
  2. The de-Broglie wavelength is associated with it.

(Take mass of alpha particle = 6.4 × 10−27 kg)


An electron is accelerated from rest through a potential difference of 100 V. Find:

  1. the wavelength associated with
  2. the momentum and
  3. the velocity required by the electron.

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.


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 ______.


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:


How will the de-Broglie wavelength associated with an electron be affected when the velocity of the electron decreases? Justify your answer.


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):


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×