This set of Class 12 Physics Chapter 11 Multiple Choice Questions & Answers (MCQs) focuses on “Wave Nature of Matter”.

1. What type of nature do electromagnetic waves have?

a) Dual nature

b) Wave nature

c) Particle nature

d) Photon nature

View Answer

Explanation: Electromagnetic radiations have a wave nature as well as properties alike to those of particles. Therefore, electromagnetic radiations are emissions with a dual nature, i.e. it has both wave and particle aspects.

2. The magnitude of which of the following is proportional to the frequency of the wave?

a) Electrons

b) Neutrons

c) Photons

d) Protons

View Answer

Explanation: The energy conveyed by an electromagnetic wave is always carried in packets whose magnitude is proportional to the frequency of the wave. These packets of energy are known as photons. The energy of a photon is given as:

E = hv

Where h is the Planck’s constant and v is the frequency of the wave.

3. Identify the de – Broglie expression from the following.

a) λ=h×p

b) λ=\(\frac {h}{p}\)

c) λ=h+p

d) λ=h-p

View Answer

Explanation: de – Broglie equation states that matter can act as waves as well as particles. So, the de Broglie equation helps us understand the concept of matter having a wavelength. The expression for de – Broglie wavelength is given as:

λ=\(\frac {h}{p} = \frac {h}{mv}\)

Where h is the Planck’s constant; p is the momentum; m is the mass of the particle and v is the velocity of the particle.

4. When the wavelength of an electron increases, the velocity of the electron will also increase.

a) True

b) False

View Answer

Explanation: No, this is a false statement. According to the de – Broglie equation, the velocity of the particle and the de – Broglie are inversely proportional to each other. Therefore, when the wavelength of an electron increases, the velocity of the electron decreases.

5. The sun gives light at the rate of 1500 W/m^{2} of area perpendicular to the direction of light. Assume the wavelength of light as 5000Å. Calculate the number of photons/s arriving at 1 m^{2} area at that part of the earth.

a) 4.770 × 10^{21}

b) 3.770 × 10^{11}

c) 3.770 × 10^{21}

d) 3.770 × 10^{20}

View Answer

Explanation: Given: I = 1500 W/m

^{2}; Wavelength = 5000Å

Required equation ➔ E=hv=\(\frac {hc}{\lambda }\)

Speed of light (c) = 3 × 10

^{8}m/s

Number of photons/s received = n = \(\frac {IA}{E} = \frac {(1500 \times 1) \times (5000 \times 10^{-10})}{6.63 \times 10^{-34} \times 3 \times 10^8} \)

n = 3.770 × 10

^{21}

Therefore, the number of electrons received per second is 3.770 × 10

^{21}.

6. What is the de – Broglie wavelength associated with an electron, accelerated through a potential difference of 200 volts?

a) 1 nm

b) 0.5 nm

c) 0.0056 nm

d) 0.086 nm

View Answer

Explanation: Given: Potential difference (V) = 200 V

The de – Broglie wavelength is given as:

λ=\(\frac {h}{p} = \frac {h}{mv} = \frac {1.227}{\sqrt {V}}\) nm

λ=\(\frac {1.227}{\sqrt {200}}\)

λ=0.086 nm

7. What is the de – Broglie wavelength of a proton accelerated through a potential difference of 2 kV?

a) 0.65 × 10^{-13} m

b) 0.65 × 10^{-10} m

c) 0.65 × 10^{-11} m

d) 0.65 × 10^{-20} m

View Answer

Explanation: Given: charge of proton = 1.6 × 10

^{-19}; mass of proton = 1.6 × 10

^{-27}; V = 2 kV

λ=\(\frac {h}{p} = \frac {h}{\sqrt {2mE}} = \frac {h}{\sqrt {2mqV}}\)

λ=\(\frac {6.6 \times 10^{-34}}{\sqrt {2\times (1.6 \times 10^{-27})\times (1.6 \times 10^{-19}) \times 2000}}\)

λ=0.65 × 10

^{-12}m

Therefore, the wavelength is 0.65 × 10

^{-12}m.

8. While comparing the alpha particle, neutron, and beta particle, the alpha particle has the lowest de – Broglie wavelength.

a) True

b) False

View Answer

Explanation: Yes, this is a true statement. In comparison with beta particle and neutron, the alpha particle has a higher mass, followed by neutron and then beta particle. According to de – Broglie wavelength equation, the wavelength is inversely proportional to the mass. Hence, the alpha particle has the lowest wavelength since it has the highest mass.

9. What is the de – Broglie wavelength of a ball of mass 150 g moving at a speed of 50 m/s?

a) 8.8 × 10^{-34} m

b) 8.8 × 10^{-30} m

c) 8.8 × 10^{-25} m

d) 8.8 × 10^{-35} m

View Answer

Explanation: Given: m = 150 g; v = 50 m/s

The required equation ➔ λ=\(\frac {h}{p} = \frac {h}{mv}\)

λ=\(\frac {6.6 \times 10^{-34}}{150 \times 10^{-3} \times 50}\)

λ=8.8 × 10

^{-35}m

10. What will be the de – Broglie wavelength when the kinetic energy of the electron increases by 5 times?

a) √5

b) 5

c) \(\frac {1}{\sqrt {5}}\)

d) \(\frac {1}{5}\)

View Answer

Explanation: The required equation ➔ λ=\(\frac {h}{mv} = \frac {h}{\sqrt {2mK}} \)

Where h is the Planck’s constant, m is the mass of the electron and K is the kinetic energy of the electron.

Since the mass of the electron remain unchanged, the wavelength will be inversely proportioned to the kinetic energy, so,

\(\frac {\lambda }{\lambda^{‘}} = \sqrt {\frac {K^{‘}}{K} } = \sqrt { \frac {5K}{K} } \) = √5

Therefore, λ’=\(\frac {\lambda }{\sqrt {5}}\)

Hence, the wavelength is reduced by a factor of √5.

**Sanfoundry Global Education & Learning Series – Physics – Class 12**.

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