Matter Waves and Davisson–Germer Experiment – Matter waves-wave nature of particle: NEET MCQs with Answers

15 free questions on Matter Waves and Davisson–Germer Experiment – Matter waves-wave nature of particle for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Matter waves-wave nature of particle, de Broglie relation.

Practise all 97 Matter Waves and Davisson–Germer Experiment – Matter waves-wave nature of particle questions free Take a free timed test

Question 1

Previous-year question
Which of the following figures represents the variation of particle momentum and the associated de Broglie wavelength? [AIPMT 2015]
Figure for question 1
  1. Graph (a)
  2. Graph (b)
  3. Graph (c)
  4. Graph (d)
Show answer
Answer: (B) Graph (b)
Explanation: From λ = h/p we have pλ = h, a constant. This is the equation of a rectangular hyperbola, so the graph of p against λ falls steeply at small λ and flattens towards the axis at large λ — graph (b).

Question 2

Previous-year question
An electron is accelerated from rest through a potential difference of V volt. If the de-Broglie wavelength of the electron is 1.227 × 10⁻² nm, the potential difference is [NEET 2020]
  1. 10² V
  2. 10³ V
  3. 10⁴ V
  4. 10 V
Show answer
Answer: (C) 10⁴ V
Explanation: λ = 1.227/√V nm. Setting 1.227/√V = 1.227 × 10⁻² gives √V = 100, so V = 10⁴ V.

Question 3

Previous-year question
An electron is accelerated through a potential difference of 10,000 V. Its de-Broglie wavelength is nearly (mass of electron = 9 × 10⁻³¹ kg) [NEET 2019]
  1. 12.2 × 10⁻¹³ m
  2. 12.2 × 10⁻¹² m
  3. 12.2 × 10⁻¹⁴ m
  4. 12.2 nm
Show answer
Answer: (B) 12.2 × 10⁻¹² m
Explanation: λ = 12.27/√V Å = 12.27/100 = 0.1227 Å = 12.27 × 10⁻¹² m.

Question 4

Previous-year question
The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (in kelvin) and having mass m is [NEET 2017]
  1. h/√(3mkT)
  2. 2h/√(3mkT)
  3. 2h/√(mkT)
  4. h/√(mkT)
Show answer
Answer: (A) h/√(3mkT)
Explanation: A thermal neutron has average kinetic energy (3/2)kT, so p = √(2m × 3kT/2) = √(3mkT) and λ = h/√(3mkT).

Question 5

Previous-year question
An electron of mass m and a photon have the same energy E. The ratio of the de-Broglie wavelength of the electron to that of the photon is [NEET 2016]
  1. (1/c)(E/2m)^½
  2. (E/2m)^½
  3. c(2mE)^(−½)
  4. (1/xc)(2m/E)^½
Show answer
Answer: (A) (1/c)(E/2m)^½
Explanation: For the electron λ(e) = h/√(2mE); for the photon λ(ph) = hc/E. The ratio is λ(e)/λ(ph) = E/(c√(2mE)) = (1/c)√(E/2m).

Question 6

Previous-year question
Light of wavelength 500 nm is incident on a metal of work function 2.28 eV. The de-Broglie wavelength of the emitted electron is [AIPMT 2015]
  1. < 2.8 × 10⁻⁹ m
  2. ≥ 2.8 × 10⁻⁹ m
  3. ≤ 2.8 × 10⁻¹² m
  4. < 2.8 × 10⁻¹⁰ m
Show answer
Answer: (B) ≥ 2.8 × 10⁻⁹ m
Explanation: Photon energy = 1240/500 = 2.48 eV, so K.E.(max) = 0.2 eV. The fastest electron has λ = 12.27/√0.2 Å ≈ 27.4 Å ≈ 2.8 × 10⁻⁹ m. Slower electrons have longer wavelengths, so λ ≥ 2.8 × 10⁻⁹ m.

Question 7

Previous-year question
An electron of mass m with an initial velocity v = v₀ î (v₀ > 0) is in an electric field E = −E₀ î (E₀ = constant > 0). If λ₀ is its initial de-Broglie wavelength, then at time t the wavelength is [NEET 2018]
  1. λ₀/[1 + (eE₀/m)(t/v₀)]
  2. λ₀[1 + eE₀t/mv₀]
  3. λ₀
  4. λ₀t
Show answer
Answer: (A) λ₀/[1 + (eE₀/m)(t/v₀)]
Explanation: The force on the electron is −eE = +eE₀ î, so the speed grows as v = v₀ + (eE₀/m)t. Then λ = h/mv = λ₀/[1 + (eE₀/m)(t/v₀)].
Doing well? Create a free account to take timed tests on this topic and see your weak areas.

Question 8

Previous-year question
The de Broglie wave corresponding to a particle of mass m and velocity v has a wavelength associated with it [AIPMT 1989]
  1. h/mv
  2. hmv
  3. mh/v
  4. m/hv
Show answer
Answer: (A) h/mv
Explanation: De Broglie's hypothesis assigns to every moving particle a wave of wavelength λ = h/p = h/mv. The wavelength is inversely proportional to both mass and velocity, which is why it is measurable for electrons but utterly negligible for macroscopic bodies.

Question 9

Previous-year question
An electron beam has a kinetic energy equal to 100 eV. Find its wavelength associated with a beam, if mass of electron = 9.1 × 10⁻³¹ kg and 1 eV = 1.6 × 10⁻¹⁹ J. (Planck's constant = 6.6 × 10⁻³⁴ J s) [AIPMT 1996]
  1. 24.6 Å
  2. 0.12 Å
  3. 1.2 Å
  4. 6.3 Å
Show answer
Answer: (C) 1.2 Å
Explanation: λ = h/√(2mE) = 6.6 × 10⁻³⁴/√(2 × 9.1 × 10⁻³¹ × 100 × 1.6 × 10⁻¹⁹) = 6.6 × 10⁻³⁴/√(2.912 × 10⁻⁴⁷). This equals 6.6 × 10⁻³⁴/5.4 × 10⁻²⁴ = 1.2 × 10⁻¹⁰ m = 1.2 Å.

Question 10

Previous-year question
The wavelength associated with an electron, accelerated through a potential difference of 100 V, is of the order of [AIPMT 1996]
  1. 1000 Å
  2. 100 Å
  3. 10.5 Å
  4. 1.2 Å
Show answer
Answer: (D) 1.2 Å
Explanation: λ = 12.27/√V Å = 12.27/√100 = 12.27/10 = 1.227 Å ≈ 1.2 Å. Wavelengths of this order are comparable with interatomic spacings, which is why such electrons are diffracted by crystals.

Question 11

Previous-year question
The energy of a photon of light is 3 eV. Then the wavelength of the photon must be [AIPMT 2000]
  1. 4125 nm
  2. 412.5 nm
  3. 41250 nm
  4. 4 nm
Show answer
Answer: (B) 412.5 nm
Explanation: Using λ(Å) = 12375/E(eV), λ = 12375/3 = 4125 Å. Converting, 4125 Å = 412.5 nm — a wavelength in the visible violet-blue region.

Question 12

Previous-year question
The following particles are moving with the same velocity, then the maximum de Broglie wavelength will be for [AIPMT 2002]
  1. proton
  2. α-particle
  3. neutron
  4. β-particle
Show answer
Answer: (D) β-particle
Explanation: At a common velocity, λ = h/mv is inversely proportional to mass. The β-particle (an electron) is by far the lightest of the four, so it has the greatest de Broglie wavelength.

Question 13

Previous-year question
A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 × 10⁶ m s⁻¹. The velocity of the particle is (mass of electron = 9.1 × 10⁻³¹ kg) [AIPMT 2008]
  1. 2.7 × 10⁻¹⁸ m s⁻¹
  2. 9 × 10⁻² m s⁻¹
  3. 3 × 10⁻³¹ m s⁻¹
  4. 2.7 × 10⁻²¹ m s⁻¹
Show answer
Answer: (A) 2.7 × 10⁻¹⁸ m s⁻¹
Explanation: Equal wavelengths mean equal momenta: mv = m_ev_e. Hence v = (9.1 × 10⁻³¹ × 3 × 10⁶)/(1 × 10⁻⁶) = 2.73 × 10⁻²⁴/10⁻⁶ = 2.7 × 10⁻¹⁸ m s⁻¹.

Question 14

Previous-year question
Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV then the de Broglie wavelength associated with the electrons would [AIPMT 2011]
  1. increase by 2 times
  2. decrease by 2 times
  3. decrease by 4 times
  4. increase by 4 times
Show answer
Answer: (B) decrease by 2 times
Explanation: Since λ = 1.227/√V nm, λ ∝ 1/√V. Then λ₁/λ₂ = √(V₂/V₁) = √(100/25) = 2, so λ₂ = λ₁/2 — the wavelength decreases by a factor of 2.

Question 15

Previous-year question
If the momentum of an electron is changed by p, then the de Broglie wavelength associated with it changes by 0.5%. The initial momentum of the electron will be [AIPMT 2012]
  1. 200p
  2. 400p
  3. p/200
  4. 100p
Show answer
Answer: (A) 200p
Explanation: Since λ = h/P, differentiating gives |Δλ/λ| = |ΔP/P|. Here ΔP = p and Δλ/λ = 0.5/100, so p/P = 0.005 and P = 200p.

Want more Matter Waves and Davisson–Germer Experiment – Matter waves-wave nature of particle questions?

E-GYAN NIDHI has 97 questions on this topic and 63,000+ in total, all free: timed CBT tests, NEET marking and instant analysis.

Register free Free Sunday Mock Test

More topics in DUAL NATURE OF MATTER AND RADIATION

↑