Physics Assessment Questions for IIT JEE Exam 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
Answer: a
Clarification: 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
Answer: c
Clarification: 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
Answer: b
Clarification: 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
Answer: b
Clarification: 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/m2 of area perpendicular to the direction of light. Assume the wavelength of light as 5000Å. Calculate the number of photons/s arriving at 1 m2 area at that part of the earth.
a) 4.770 × 1021
b) 3.770 × 1011
c) 3.770 × 1021
d) 3.770 × 1020
Answer: c
Clarification: Given: I = 1500 W/m2; Wavelength = 5000Å
Required equation ➔ E=hv=(frac {hc}{lambda })
Speed of light (c) = 3 × 108 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 × 1021
Therefore, the number of electrons received per second is 3.770 × 1021.
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
Answer: d
Clarification: 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