Physics - Dual Nature Radiation Matter - Quiz-2

Cathode rays are similar to visible light rays in that

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They both can be deflected by electric and magnetic fields
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They both have a definite magnitude of wavelength
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They both can ionize a gas through which they pass
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They both can expose a photographic plate

Electron volt is a unit of

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Potential
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Charge
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Power
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Energy

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

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$\frac{2 eV}{m}$
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$\sqrt{\frac{2 eV}{m}}$
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$\sqrt{\frac{2 m}{eV}}$
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$\frac{V^{2}}{2 em}$

The idea of matter waves was given by

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Davisson and Germer
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de-Broglie
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Einstein
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Planck

Wave is associated with matter:

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when it is stationary.
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when it is in motion with the velocity of light only.
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when it is in motion with any velocity.
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None of the above.

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

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h/mv
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mv/h
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mh/v
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m/hv

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

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Is the photon
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Is the electron
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Is the uranium nucleus
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Depends upon the wavelength and the properties of the particle

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

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equal to c, the speed of light in vacuum.
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greater than c.
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less than c.
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tending to infinity.

When the kinetic energy of an electron is increased, the wavelength of the associated wave will

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Increase
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Decrease
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Wavelength does not depend on the kinetic energy
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None of the above

If the de-Broglie wavelengths for a proton and for an $α$ -particle are equal, then the ratio of their velocities will be:

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4 : 1
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2 : 1
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1 : 2
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1 : 4

The de-Broglie wavelength $λ$ associated with an electron having kinetic energy E is given by the expression

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$\frac{h}{\sqrt{2 mE}}$
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$\frac{2 h}{mE}$
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2mhE
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$\frac{2 \sqrt{2 mE}}{h}$

Dual nature of radiation is shown by:

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Diffraction and reflection
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Refraction and diffraction
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Photoelectric effect alone
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Photoelectric effect and diffraction

The momentum of a photon in an X-ray beam of $10^{- 10}$ metre wavelength is

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$1.5 \times 10^{- 23}$ kg-m/sec
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$6.6 \times 10^{- 24}$ kg-m/sec
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$6.6 \times 10^{- 44}$ kg-m/sec
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$2.2 \times 10^{- 52}$ kg-m/sec

An electron of mass m when accelerated through a potential difference V has de-Broglie wavelength $λ$ . The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be

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$λ \frac{m}{M}$
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$λ \sqrt{\frac{m}{M}}$
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$λ \frac{M}{m}$
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$λ \sqrt{\frac{M}{m}}$

What is the de-Broglie wavelength of the $α$ -particle accelerated through a potential difference V

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$\frac{0.287}{\sqrt{V}}$ Å
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$\frac{12.27}{\sqrt{V}}$ Å
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$\frac{0.101}{\sqrt{V}}$ Å
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$\frac{0.202}{\sqrt{V}}$ Å

The energy that should be added to an electron, to reduce its de-Broglie wavelengths from $10^{- 10}$ m to $0.5 \times 10^{- 10}$ m, will be:

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four times the initial energy.
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thrice the initial energy.
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equal to the initial energy.
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twice the initial energy.

The de-Broglie wavelength of an electron having 80eV of energy is nearly
(1eV = $1.6 \times 10^{- 19}$ J, Mass of electron = $9 \times 10^{- 31}$ Kg Plank’s constant = $6.6 \times 10^{- 34}$ J-sec)

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() 140 Å
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() 0.14 Å
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() 14 Å
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() 1.4 Å

If particles are moving with same velocity, then maximum de-Broglie wavelength will be for

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Neutron
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Proton
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$β$ -particle
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$α$ -particle

If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same

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Energy
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Momentum
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Velocity
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Angular momentum

The de-Broglie wavelength is proportional to

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$λ \propto \frac{1}{v}$
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$λ \propto \frac{1}{m}$
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$λ \propto \frac{1}{p}$
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$λ \propto p$

Particle nature and wave nature of electromagnetic waves and electrons can be shown by

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Electron has small mass, deflected by the metal sheet
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X-ray is diffracted, reflected by thick metal sheet
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Light is refracted and defracted
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Photoelectricity and electron microscopy

The de-Broglie wavelength of a particle moving with a velocity $2.25 \times 10^{8}$ m/s is equal to the wavelength of the photon. The ratio of the kinetic energy of the particle to the energy of the photon is (velocity of light is $3 \times 10^{8}$ m/s)

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1/8
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3/8
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5/8
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7/8

The speed of an electron having a wavelength of $10^{- 10}$ m is

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() $7.25 \times 10^{6}$ m/s
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() $6.26 \times 10^{6}$ m/s
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() $5.25 \times 10^{6}$ m/s
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() $4.24 \times 10^{6}$ m/s

The kinetic energy of electron and proton is $10^{- 32}$ J. Then the relation between their de-Broglie wavelengths is

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$λ_{p} < λ_{e}$
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$λ_{p} > λ_{e}$
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$λ_{p} = λ_{e}$
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$λ_{p} = 2 λ_{e}$

The de-Broglie wavelength of a particle accelerated with 150 volt potential is $10^{- 10}$ m. If it is accelerated by 600 volts p.d., its wavelength will be

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0.25 Å
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0.5 Å
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1.5 Å
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2 Å

The de-Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of 3 km/s will be

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1 Å
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0.66 Å
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6.6 Å
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66 Å

When the momentum of a proton is changed by an amount $P_{0}$ , the corresponding change in the de-Broglie wavelength is found to be 0.25%. Then, the original momentum of the proton was

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() $P_{0}$
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() 100 $P_{0}$
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() 400 $P_{0}$
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() 4 $P_{0}$

The de-Broglie wavelength of a neutron at 27 $° C$ is $λ$ . What will be its wavelength at 927 $° C$

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() $λ$ / 2
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() $λ$ / 3
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() $λ$ / 4
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() $λ$ / 9

An electron and proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is

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Zero
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Infinity
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Equal to the kinetic energy of the proton
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Greater than the kinetic energy of the proton

For moving ball of cricket, the correct statement about de-Broglie wavelength is

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It is not applicable for such big particle
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$\frac{h}{\sqrt{2 mE}}$
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$\sqrt{\frac{h}{2 mE}}$
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$\frac{h}{2 mE}$

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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