Class 12 Physics Chapter 11: Dual Nature of Radiation and Matter worksheet with answer key

Answer: (b) are none at all

Emission needs hν > φ₀, that is ν > ν₀. Below ν₀ no electron is emitted, however intense the radiation or however long it falls, because each electron absorbs a single quantum, and one quantum is too small.

Answer: (a) A photon is deflected by a magnetic field.

Photons are electrically neutral and are not deflected by electric or magnetic fields. The other three statements are photon properties: E = hν, speed c, and in a photon-particle collision the number of photons may not be conserved.

Answer: (b) becomes positively charged

Ultraviolet light ejects negatively charged particles (electrons) from zinc, so an initially uncharged plate is left positively charged, as Hallwachs observed.

Answer: (a) within about 10⁻⁹ s or less

Above the threshold frequency emission starts without apparent time lag even for very dim radiation, in a time of the order of 10⁻⁹ s or less. A delay that grows as the intensity falls is what the wave picture predicts, and is not observed.

Answer: (d) the stopping potential is larger in magnitude and the saturation current is the same

At the same intensity, curves for different frequencies reach the same saturation current, but the stopping potential is more negative for the higher frequency: the greater the frequency, the greater Kmax and hence V₀.

Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(300 × 10⁻⁹) = 6.63 × 10⁻¹⁹ J = 6.63 × 10⁻¹⁹/1.6 × 10⁻¹⁹ ≈ 4.1 eV. Emission needs E > φ₀, so only X emits, with Kmax = 4.1 − 2.3 ≈ 1.8 eV; Y (4.5 eV) does not. For Y, λ₀ = hc/φ₀ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(4.5 × 1.6 × 10⁻¹⁹) ≈ 2.8 × 10⁻⁷ m = 276 nm; only light shorter than this ejects electrons from Y.

• E ≈ 4.1 eV; only X emits, Kmax ≈ 1.8 eV (1)
• λ₀ for Y ≈ 276 nm (1)

In the wave picture the energy of the radiation is spread continuously over the whole wavefront and shared by a very large number of electrons, so each absorbs only a little per unit time; calculations estimate that an electron could take hours or more to gather the work function. Observed: above the threshold, emission starts instantaneously, in about 10⁻⁹ s or less, even for very dim light.

• wave picture: energy shared over the wavefront, time lag of hours or more (1)
• observed: no time lag, ~10⁻⁹ s or less (1)

Number per second N = P/E = Pλ/hc. At equal power N ∝ λ, so the red lamp, whose photons each carry less energy, emits more photons per second. N(red)/N(violet) = 700/400 = 1.75.

• N = P/E = Pλ/hc, so red emits more (1)
• ratio 1.75 (1)

Kmax = hν − φ₀, where an electron absorbs one quantum hν and spends at least φ₀ to escape. (i) Kmax cannot be negative, so emission needs hν > φ₀, that is ν > ν₀ = φ₀/h; below ν₀ no single quantum has enough energy, however many quanta arrive. (ii) Intensity fixes the number of photons per second, not the energy of each; since one quantum is absorbed by one electron, Kmax is fixed by hν and φ₀ alone.

• Kmax = hν − φ₀ with terms explained (1)
• threshold: hν > φ₀ needed, ν₀ = φ₀/h (1)
• intensity = photon number; one photon per electron (1)

eV₀ = hν − φ₀. In eV: hν₁ = 1.0 + φ₀ and 1.5hν₁ = 2.5 + φ₀. Substituting, 1.5(1.0 + φ₀) = 2.5 + φ₀, so 0.5φ₀ = 1.0 and (a) φ₀ = 2.0 eV. (b) hν₁ = 3.0 eV, so ν₁ = 3.0 × 1.6 × 10⁻¹⁹/6.63 × 10⁻³⁴ ≈ 7.2 × 10¹⁴ Hz. (c) ν₀ = φ₀/h = 2.0 × 1.6 × 10⁻¹⁹/6.63 × 10⁻³⁴ ≈ 4.8 × 10¹⁴ Hz.

• two equations, φ₀ = 2.0 eV (1)
• ν₁ ≈ 7.2 × 10¹⁴ Hz (1)
• ν₀ ≈ 4.8 × 10¹⁴ Hz (1)

(a) Using mutually perpendicular electric and magnetic fields, Thomson measured the speed and the specific charge e/m of cathode ray particles. e/m was found to be the same whatever the cathode metal or the gas in the tube, which suggested that the particles are universal constituents of matter (electrons). (b) m = e/(e/m) = 1.602 × 10⁻¹⁹/1.76 × 10¹¹ ≈ 9.1 × 10⁻³¹ kg.

• speed and e/m measured with crossed fields (1)
• e/m independent of metal and gas, so universal (1)
• m ≈ 9.1 × 10⁻³¹ kg (1)