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प्रश्न
Describe briefly how the Davisson-Germer experiment demonstrated the wave nature of electrons.
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उत्तर
Experiment:
Experimental arrangement used by Davisson and Germer: Electrons from a hot tungsten cathode are accelerated by a potential difference V between the cathode (C) and anode (A).
A narrow hole in the anode renders the electrons into a fine beam of electrons and allows them to strike a nickel crystal.
The electrons are scattered in all directions by the atoms in the crystal and its intensity in a given direction is found by the use of a detector.
The graph is plotted between angle Φ (angle between incident and the scattered direction of the electron beam) and intensity of the scattered beam.
The experimental curves obtained by Davisson and Germer are as shown in the figure below:
संबंधित प्रश्न
Describe the construction of photoelectric cell.
What is the de Broglie wavelength of a ball of mass 0.060 kg moving at a speed of 1.0 m/s?
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)
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`
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.
An alpha particle is accelerated through a potential difference of 100 V. Calculate:
- The speed acquired by the alpha particle, and
- The de-Broglie wavelength is associated with it.
(Take mass of alpha particle = 6.4 × 10−27 kg)
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 ______.
For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?
Matter waves are ______.
