Q: The de Broglie wavelength of a cricket ball of mass 0.1 kg moving at 10 m/s (h = 6.6 × 10⁻³⁴ J s) is about:

6.6 \times 10^{-34} m. λ = h/mv = (6.6 × 10⁻³⁴)/(0.1 × 10) = 6.6 × 10⁻³⁴ m — far too small to detect, so the wave nature of large objects is not noticeable.

Question 1

de Broglie proposed that matter (like electrons) has:

Accepted Answer

A dual particle and wave nature. de Broglie suggested that, like light, all moving matter has both particle and wave character.

Question 2

The de Broglie wavelength (λ) of a particle of mass m moving with velocity v is:

Accepted Answer

λ = h/mv. The de Broglie relation is λ = h/p = h/mv, where h is Planck's constant.

Question 3

The de Broglie wavelength of a particle is inversely proportional to its:

Accepted Answer

Momentum. Since λ = h/p, wavelength is inversely proportional to momentum p = mv.

Question 4

de Broglie (matter) waves are significant (observable) only for:

Accepted Answer

Microscopic particles of very small mass. Because h is extremely small, an appreciable λ requires a very small mass — so wave behaviour is only significant for microscopic particles like electrons.

Question 5

In terms of kinetic energy (KE), the de Broglie wavelength of a particle is:

Accepted Answer

λ = h/√(2m·KE). Since KE = p²/2m, we get p = √(2m·KE), so λ = h/p = h/√(2m·KE).

Question 6

If an electron, a proton, an α-particle and a neutron all have the same kinetic energy, which has the LONGEST de Broglie wavelength?

Accepted Answer

Electron. λ = h/√(2m·KE); for equal KE, the smallest mass gives the largest λ — that is the electron.

Question 7

The de Broglie wavelength of an electron accelerated through a potential difference of 100 V is (λ = 12.27/√V Å):

Accepted Answer

1.227 Å. λ = 12.27/√V Å = 12.27/√100 = 12.27/10 = 1.227 Å.

Question 8

The wave nature of the electron was experimentally confirmed by the electron-diffraction experiment of:

Accepted Answer

Davisson and Germer. Davisson and Germer observed diffraction of electrons by a nickel crystal, confirming de Broglie's hypothesis.

Question 9

The de Broglie wavelength of a cricket ball of mass 0.1 kg moving at 10 m/s (h = 6.6 × 10⁻³⁴ J s) is about:

Accepted Answer

6.6 	imes 10^{-34} m. λ = h/mv = (6.6 × 10⁻³⁴)/(0.1 × 10) = 6.6 × 10⁻³⁴ m — far too small to detect, so the wave nature of large objects is not noticeable.

Question 10

If the velocity of a moving particle is doubled, its de Broglie wavelength becomes:

Accepted Answer

Half. λ = h/mv is inversely proportional to velocity, so doubling v halves λ.

Question 11

A proton and an electron move with the same velocity. Which has the SHORTER de Broglie wavelength?

Accepted Answer

Proton. For equal velocity, λ = h/mv is smaller for the larger mass; the proton is much heavier, so it has the shorter wavelength.

Question 12

The de Broglie wavelength of an electron accelerated through 150 V (λ = 12.27/√V Å) is approximately:

Accepted Answer

1.0 Å. λ = 12.27/√150 ≈ 12.27/12.25 ≈ 1.0 Å.

Question 13

Heisenberg's uncertainty principle states that it is impossible to determine simultaneously and exactly the:

Accepted Answer

Position and momentum of a microscopic particle. The principle says the position and momentum (velocity) of a small particle cannot both be measured exactly at the same time.

Question 14

The mathematical form of Heisenberg's uncertainty principle is:

Accepted Answer

Δx · Δp ≥ h/4π. The product of the uncertainties in position (Δx) and momentum (Δp) is at least h/4π.

Question 15

In terms of the uncertainty in velocity (Δv), the uncertainty principle can be written as:

Accepted Answer

Δx · Δv ≥ h/4πm. Since Δp = mΔv, dividing Δx·Δp ≥ h/4π by m gives Δx·Δv ≥ h/(4πm).

Question 16

Heisenberg's uncertainty principle is significant for:

Accepted Answer

Microscopic particles such as electrons. The uncertainty is only appreciable for particles of very small mass; for everyday objects it is far too small to matter.

Question 17

The uncertainty principle is negligible for macroscopic objects because:

Accepted Answer

Their large mass makes h/4πm extremely small. For a large mass m, the quantity h/(4πm) is vanishingly small, so the uncertainties are undetectable.

Question 18

According to the uncertainty principle, if the uncertainty in the position of a particle decreases, the uncertainty in its momentum:

Accepted Answer

Increases. Since Δx·Δp ≥ h/4π is a constant lower bound, making Δx smaller forces Δp to become larger.

Question 19

The uncertainty in the position of an electron is 1 × 10⁻¹⁰ m. The minimum uncertainty in its momentum (h = 6.6 × 10⁻³⁴ J s) is about:

Accepted Answer

5.3 	imes 10^{-25} kg m/s. Δp ≥ h/(4π·Δx) = (6.6 × 10⁻³⁴)/(4 × 3.14 × 1 × 10⁻¹⁰) ≈ 5.3 × 10⁻²⁵ kg m/s.

Question 20

Heisenberg's uncertainty principle led scientists to reject the idea of:

Accepted Answer

Well-defined, fixed orbits for electrons. Because an electron's exact path cannot be known, Bohr's fixed orbits were replaced by the concept of orbitals (regions of probability).

Towards the Quantum Mechanical Model — NEET Chemistry MCQs with Solutions

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