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Dual Nature of Radiation and Matter

Telangana Open School (TOSS) · Class 12 · Physics

Flashcards for Dual Nature of Radiation and Matter — Telangana Open School (TOSS) Class 12 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.

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Card 1Photoelectric Effect

What is the photoelectric effect?

Answer

The emission of electrons from a metal surface when light of sufficiently high frequency (above threshold frequency) is incident on it. These emitted electrons are called photoelectrons.

Card 2Photoelectric Effect

Define work function of a metal.

Answer

The minimum energy required by an electron to escape from the metal surface. It is denoted by φ₀ and measured in electron volts (eV).

Card 3Photoelectric Effect

What is threshold frequency?

Answer

The minimum frequency of incident light below which no photoelectrons are emitted from a metal surface. It is denoted by ν₀ and depends on the material.

Card 4Photoelectric Effect

Why does increasing the intensity of light not increase the kinetic energy of photoelectrons?

Answer

Because the energy of each photon depends only on frequency (E = hν), not intensity. Increasing intensity increases the number of photons, hence more photoelectrons, but their maximum kinetic energy r

Card 5Photoelectric Effect

What is stopping potential?

Answer

The minimum retarding potential applied to stop the fastest photoelectrons, making the photoelectric current zero. It is related to maximum kinetic energy: eV₀ = ½mv_max².

Card 6Photoelectric Effect

A photon of frequency 1.2 × 10¹⁵ Hz strikes a metal with work function 4.3 eV. Calculate the maximum kinetic energy of the emitted electron.

Answer

Step 1: Energy of photon E = hν = (6.6 × 10⁻³⁴) × (1.2 × 10¹⁵) = 7.92 × 10⁻¹⁹ J Step 2: Convert to eV: 7.92 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 4.95 eV Step 3: K_max = E - φ₀ = 4.95 - 4.3 = 0.65 eV Answer: 0.65 e

Card 7Photoelectric Effect

Light of wavelength 400 nm is incident on sodium (work function = 2.3 eV). Will photoelectrons be emitted?

Answer

Step 1: ν = c/λ = (3 × 10⁸) / (400 × 10⁻⁹) = 7.5 × 10¹⁴ Hz Step 2: E = hν = (6.6 × 10⁻³⁴)(7.5 × 10¹⁴) = 4.95 × 10⁻¹⁹ J = 3.09 eV Step 3: Since 3.09 eV > 2.3 eV, photoelectrons will be emitted. Answer:

Card 8Photoelectric Effect

State Einstein’s photoelectric equation.

Answer

hν = φ₀ + K_max Where hν is energy of incident photon, φ₀ is work function, and K_max is maximum kinetic energy of emitted photoelectron.

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