In an electron gun, the electrons are accelerated by the potential V. If e is the charge and m is the mass of an electron, then the maximum velocity of these electrons will be

(a) $\frac{2\mathrm{eV}}{\mathrm{m}}$                    (b) $\sqrt{\frac{2\mathrm{eV}}{\mathrm{m}}}$

(c) $\sqrt{\frac{2\mathrm{m}}{\mathrm{eV}}}$                  (d) $\frac{{\mathrm{V}}^{2}}{2\mathrm{em}}$

Concept Questions :-

Electron emission
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The idea of matter waves was given by
(a) Davisson and Germer                  (b) de-Broglie
(c) Einstein                                      (d) Planck

Concept Questions :-

De-broglie wavelength
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Wave is associated with matter

(a) When it is stationary
(b) When it is in motion with the velocity of light only
(c) When it is in motion with any velocity
(d) None of the above

Concept Questions :-

De-broglie wavelength
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The de-Broglie wavelength associated with the particle of mass m moving with velocity v is

(a) h/mv                 (b) mv/h
(c) mh/v                 (d) m/hv

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A photon, an electron, and a uranium nucleus all have the same wavelength. The one with the most energy:

(a) Is the photon
(b) Is the electron
(c) Is the uranium nucleus
(d) Depends upon the wavelength and the properties of the particle

Concept Questions :-

De-broglie wavelength
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A particle which has zero rest mass and non-zero energy and momentum must travel with a speed:

(a) Equal to c, the speed of light in vacuum

(b) Greater than c

(c) Less than c

(d) Tending to infinity

Concept Questions :-

Particle nature of light
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When the kinetic energy of an electron is increased, the wavelength of the associated wave will
(a) Increase
(b) Decrease
(c) Wavelength does not depend on the kinetic energy
(d) None of the above

Concept Questions :-

De-broglie wavelength
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If the de-Broglie wavelengths for a proton and for a $\mathrm{\alpha }$-particle are equal, then the ratio of their velocities will be
(a) 4 : 1                           (b) 2 : 1
(c) 1 : 2                           (d) 1 : 4

Concept Questions :-

De-broglie wavelength
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The de-Broglie wavelength $\mathrm{\lambda }$ associated with an electron having kinetic energy E is given by the expression

(a) $\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}$                           (b) $\frac{2\mathrm{h}}{\mathrm{mE}}$

(c) 2mhE                               (d) $\frac{2\sqrt{2\mathrm{mE}}}{\mathrm{h}}$

Concept Questions :-

De-broglie wavelength
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Dual nature of radiation is shown by:

(a) Diffraction and reflection

(b) Refraction and diffraction

(c) Photoelectric effect alone

(d) Photoelectric effect and diffraction

Concept Questions :-

Photoelectric effect experiment
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