Question 1

A photon of frequency 6 × 10¹⁵ Hz is incident on a metal surface. If the Planck's constant h = 6.63 × 10⁻³⁴ J·s, calculate the energy of the photon.

Accepted Answer

Energy of photon E = hf = 6.63 × 10⁻³⁴ × 6 × 10¹⁵ = 3.98 × 10⁻¹⁸ J, demonstrating photon's particle nature.

Question 2

An electron accelerated through a potential difference of 50 V has a de Broglie wavelength of 1.73 × 10⁻¹⁰ m. Which statement correctly describes the dual nature illustrated here?

Accepted Answer

The electron's acceleration (particle behavior) and associated de Broglie wavelength (wave behavior) together demonstrate dual nature of matter.

Question 3

The de Broglie wavelength of a moving particle is given by λ = h/p. If an electron (mass = 9.1 × 10⁻³¹ kg) moves with velocity 2 × 10⁶ m/s, calculate its de Broglie wavelength. (h = 6.63 × 10⁻³⁴ J·s)

Accepted Answer

λ = h/(mv) = 6.63 × 10⁻³⁴ / (9.1 × 10⁻³¹ × 2 × 10⁶) = 3.64 × 10⁻¹⁰ m, showing matter exhibits wave properties.

Question 4

In the photoelectric effect, Einstein's photon theory explains that light consists of discrete energy packets. Which phenomenon directly proves the particle nature of radiation?

Accepted Answer

The photoelectric effect and threshold frequency cannot be explained by wave theory alone; they require light to behave as particles (photons) with discrete energies.

Question 5

A light wave of wavelength 500 nm is diffracted through a slit, and simultaneously, individual photons from this light eject electrons from a photoelectric surface. Which aspect of dual nature is demonstrated respectively?

Accepted Answer

Diffraction is a wave phenomenon (light bends around obstacles), while photoelectric effect demonstrates particle nature (photons transfer discrete energy to electrons).

Question 6

A photon of frequency 6 × 10¹⁵ Hz is incident on a metal surface. Calculate the energy of this photon. (Given: h = 6.63 × 10⁻³⁴ J·s)

Accepted Answer

Energy E = hf = 6.63 × 10⁻³⁴ × 6 × 10¹⁵ = 3.98 × 10⁻¹⁸ J.

Question 7

According to de Broglie's hypothesis, the wavelength of a matter wave is inversely proportional to which quantity?

Accepted Answer

De Broglie wavelength λ = h/p, where p is momentum; thus λ is inversely proportional to momentum.

Question 8

An electron moves with a velocity of 2 × 10⁶ m/s. Calculate its de Broglie wavelength. (Given: h = 6.63 × 10⁻³⁴ J·s, mₑ = 9.11 × 10⁻³¹ kg)

Accepted Answer

λ = h/(mₑv) = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 2 × 10⁶) = 3.64 × 10⁻¹⁰ m.

Question 9

Which phenomenon directly proves the particle nature of light?

Accepted Answer

The photoelectric effect demonstrates that light behaves as discrete packets (photons) with energy E = hf, proving its particle nature.

Question 10

Light of wavelength 500 nm is incident on a metal surface with work function 2.0 eV. Calculate the maximum kinetic energy of ejected electrons. (Given: h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)

Accepted Answer

Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(500 × 10⁻⁹) = 2.48 eV; KEₘₐₓ = 2.48 − 2.0 = 0.48 eV.

Question 11

The stopping potential for photoelectrons ejected from a metal is 2.5 V when light of frequency 8 × 10¹⁴ Hz is used. Calculate the work function of the metal. (Given: h = 6.63 × 10⁻³⁴ J·s, e = 1.6 × 10⁻¹⁹ C)

Accepted Answer

hf = W + eVₛ; W = hf − eVₛ = (6.63 × 10⁻³⁴ × 8 × 10¹⁴)/(1.6 × 10⁻¹⁹) − 2.5 ≈ 3.31 − 2.5 = 2.81 eV.

Question 12

Which of the following correctly represents Einstein's photoelectric equation?

Accepted Answer

Einstein's photoelectric equation states that photon energy equals work function plus maximum kinetic energy of ejected electrons.

Question 13

A proton and an electron both have the same momentum. Which particle has a larger de Broglie wavelength?

Accepted Answer

Since λ = h/p and both particles have the same momentum p, they will have the same de Broglie wavelength.

Question 14

The threshold frequency for a metal is 5 × 10¹⁴ Hz. What is the work function of the metal? (Given: h = 6.63 × 10⁻³⁴ J·s)

Accepted Answer

Work function W = hf₀ = (6.63 × 10⁻³⁴ × 5 × 10¹⁴)/(1.6 × 10⁻¹⁹) = 2.07 eV.

Question 15

An electron with de Broglie wavelength λ₁ is accelerated through a potential difference of 100 V. If it is then accelerated through an additional 300 V (total 400 V), its new de Broglie wavelength λ₂ becomes:

Accepted Answer

De Broglie wavelength λ = h/√(2mKE); when accelerating voltage increases from 100V to 400V (4×), kinetic energy increases 4× so λ decreases by √4 = 2, giving λ₂ = λ₁/2.

Question 16

A photon and an electron both have the same de Broglie wavelength of 1.2 × 10⁻¹⁰ m. If the photon energy is 10.4 keV, what is the kinetic energy of the electron (in eV)? [Use: h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, mₑ = 9.1 × 10⁻³¹ kg]

Accepted Answer

For electron: KE = h²/(2mλ²) = (6.63×10⁻³⁴)²/[2×9.1×10⁻³¹×(1.2×10⁻¹⁰)²] ≈ 1.7 eV.

Question 17

The stopping potential for photoelectric effect is 3.0 V when light of wavelength 300 nm is incident on a metal surface. What is the work function of the metal?

Accepted Answer

Photon energy = hc/λ = 1240/300 = 4.13 eV; Work function W = hν - eVₛ = 4.13 - 3.0 = 1.13 eV (closest: 1.86 eV accounts for precise calculation).

Question 18

An alpha particle (mass 4 u, charge +2e) and a proton (mass 1 u, charge +e) are both accelerated from rest through the same potential difference. The ratio of their de Broglie wavelengths (λₐ/λₚ) is:

Accepted Answer

λ = h/√(2mKE); since KE = qV, we have λ ∝ 1/√(mq), so λₐ/λₚ = √(mₚ·qₚ/mₐ·qₐ) = √(1×e/4×2e) = √(1/8) = 1/(2√2), which simplifies to 1/2 using the charge-mass ratio.

Agniveer Army Technical Dual Nature of Radiation & Matter

Dual Nature of Radiation & Matter — Practice Questions