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Dual Nature of Matter and Radiation MCQs for NEET — Physics Questions with Answers

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The idea of matter waves was given by

Davisson and Germer

de-Broglie

Einstein

Planck

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Explanation

(b) The idea of matter waves was given by de-Broglie.

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Wave is associated with matter

When it is stationary

When it is in motion with the velocity of light only

When it is in motion with any velocity

None of the above

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Explanation

(c)

According to the de-Broglie hypothesis, a wave is associated with the matter when it is in motion i.e. it is in motion with a velocity.

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

h/mv

mv/h

mh/v

m/hv

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Explanation

(a) $λ = \frac{h}{p} = \frac{h}{mv}$

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

Is the photon

Is the electron

Is the uranium nucleus

Depends upon the wavelength and the properties of the particle

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Explanation

(a)

$λ = \frac{h}{mv} = \frac{h}{\sqrt{2 mE}} ∴ E = \frac{h^{2}}{2 {mλ}^{2}}$

$λ$ is same for all, so $E \propto \frac{1}{m}$ . Hence energy will be maximum for particle with lesser mass.

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

Equal to c, the speed of light in vacuum

Greater than c

Less than c

Tending to infinity

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Explanation

(a) Particle is photon and it travels with the velocity equal to light in vacuum.

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

Increase

Decrease

Wavelength does not depend on the kinetic energy

None of the above

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Explanation

(b) $λ = \frac{h}{p} = \frac{h}{\sqrt{2 mE}}; ∴ λ \propto \frac{1}{\sqrt{E}}$ ( h and m = constant)

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

4 : 1

2 : 1

1 : 2

1 : 4

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Explanation

(a) $λ = \frac{h}{m_{1} v_{1}} = \frac{h}{m_{2} v_{2}} ∴ \frac{v_{1}}{v_{2}} = \frac{m_{2}}{m_{1}} = \frac{4}{1}$

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

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

$\frac{2 h}{mE}$

2mhE

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

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Explanation

(a) $\frac{1}{2} {mv}^{2} = E \Rightarrow mv = \sqrt{2 mE}; ∴ λ = \frac{h}{mv} = \frac{h}{\sqrt{2 mE}}$

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Dual nature of radiation is shown by:

Diffraction and reflection

Refraction and diffraction

Photoelectric effect alone

Photoelectric effect and diffraction

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Explanation

(d) $\{Photoelectric effect \to Particle nature Diffraction \to Wave nature\} Dual nature$

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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

$λ \frac{m}{M}$

$λ \sqrt{\frac{m}{M}}$

$λ \frac{M}{m}$

$λ \sqrt{\frac{M}{m}}$

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Explanation

(b) $λ = \frac{h}{\sqrt{2 mE}} \Rightarrow λ \propto \frac{1}{\sqrt{m}}$ ( E = same)

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