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Physics MCQs for NEET — Practice Questions with Answers

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

$\frac{0.287}{\sqrt{V}}$ Å

$\frac{12.27}{\sqrt{V}}$ Å

$\frac{0.101}{\sqrt{V}}$ Å

$\frac{0.202}{\sqrt{V}}$ Å

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Explanation

On putting $Q_{α} = 2 \times 1.6 \times 10^{- 19} C$

$m_{α} = 4 m_{P} = 4 \times 1.67 \times 10^{- 27} kg \Rightarrow λ = \frac{0.101}{\sqrt{V}}$ Å

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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:

Four times the initial energy

Thrice the initial energy

Equal to the initial energy

Twice the initial energy

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Explanation

(b)

$λ = \frac{h}{\sqrt{2 mE}} \Rightarrow λ \propto \frac{1}{\sqrt{E}} \Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{E_{2}}{E_{1}}} \Rightarrow \frac{10^{- 10}}{0.5 \times 10^{- 10}} = \sqrt{\frac{E_{2}}{E_{1}}} \Rightarrow E_{2} = 4 E_{1}$

Hence added energy = $E_{2} - E_{1} = 3 E_{1}$

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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)

140 Å

0.14 Å

14 Å

1.4 Å

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Explanation

(d) $λ = \frac{h}{\sqrt{2 mE}} = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9 \times 10^{- 31} \times 80 \times 1.6 \times 10^{- 19}}} = 1.4$ Å

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

Neutron

Proton

$β$ -particle

$α$ -particle

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Explanation

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

Energy

Momentum

Velocity

Angular momentum

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Explanation

(b) If an electron and a photon propagates in the form of waves having the same wavelength, it implies that they have same momentum. This is according to de-Broglie equation, $p \propto \frac{1}{λ}$

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The de-Broglie wavelength is proportional to

$λ \propto \frac{1}{v}$

$λ \propto \frac{1}{m}$

$λ \propto \frac{1}{p}$

$λ \propto p$

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Explanation

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

Electron has small mass, deflected by the metal sheet

X-ray is diffracted, reflected by thick metal sheet

Light is refracted and defracted

Photoelectricity and electron microscopy

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Explanation

(d) In photoelectric effect particle nature of electron is shown. While in electron microscope, beam of electron is considered as electron wave.

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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)

1/8

3/8

5/8

7/8

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Explanation

(b)

$K_{particle} = \frac{1}{2} {mv}^{2} also λ = \frac{h}{mv} \Rightarrow K_{particle} = \frac{1}{2} (\frac{h}{λv}) . v^{2} = \frac{vh}{2 λ} . . . . (i) K_{photon} = \frac{hc}{λ} . . . . (ii) ∴ \frac{K_{particle}}{K_{photon}} = \frac{v}{2 c} = \frac{2.25 \times 10^{8}}{2 \times 3 \times 10^{8}} = \frac{3}{8}$

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