More practice questions

4. Q4. Find the energy of a photon (i) of frequency 3 × 10^¹⁵ Hz; (ii) of wavelength 0.50 Å (NCERT Exercise 2.6, page 69).

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   (i) E = hν = 6.626 × 10^⁻³⁴ × 3 × 10^¹⁵ = 1.988 × 10^⁻¹⁸ J. (ii) E = hc/λ = (6.626 × 10^⁻³⁴ × 3.0 × 10^⁸) ÷ (0.50 × 10^⁻¹⁰) = 3.98 × 10^⁻¹⁵ J.

5. Q5. Radiation of wavelength 242 nm is just enough to ionise a sodium atom. Find the ionisation energy of sodium in kJ mol^⁻¹ (NCERT Exercise 2.10, page 69).

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   E = hc/λ = (6.626 × 10^⁻³⁴ × 3.0 × 10^⁸) ÷ (242 × 10^⁻⁹) = 8.21 × 10^⁻¹⁹ J per atom; × 6.022 × 10^²³ mol^⁻¹ = 4.95 × 10^⁵ J mol^⁻¹ ≈ 495 kJ mol^⁻¹.

6. Q6. An ideal source emits 25 W of monochromatic yellow light of wavelength 0.57 µm (emitted optical power). How many quanta does it emit per second (NCERT Exercise 2.11, page 69)?

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   E = (6.626 × 10^⁻³⁴ × 3.0 × 10^⁸) ÷ (0.57 × 10^⁻⁶) = 3.49 × 10^⁻¹⁹ J. Quanta per second = (25 J s^⁻¹) ÷ (3.49 × 10^⁻¹⁹ J) ≈ 7.17 × 10^¹⁹ s^⁻¹.

7. Q7. Arrange in increasing order of frequency: (a) a microwave oven’s radiation; (b) amber light from a traffic signal; (c) radiation from FM radio; (d) cosmic rays; (e) X-rays (NCERT Exercise 2.45, page 71).

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   For the electromagnetic entries: (c) FM radio < (a) microwaves < (b) amber light < (e) X-rays. Cosmic rays cannot be assigned one position on this electromagnetic-frequency scale: they are mainly energetic material particles, especially protons and nuclei (NASA, Cosmic Rays; CERN, Cosmic rays: particles from outer space). NCERT’s item uses the older label as if it meant extremely energetic electromagnetic radiation; if gamma radiation were intended, it would conventionally follow X-rays, though their energy ranges overlap.

8. Q8. The work function of caesium is 1.9 eV. Find (a) its threshold wavelength and (b) its threshold frequency. If caesium is irradiated with light of wavelength 500 nm, find the kinetic energy and speed of the ejected electrons (NCERT Exercise 2.51, page 72).

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   W₀ = 1.9 × 1.602 × 10^⁻¹⁹ = 3.044 × 10^⁻¹⁹ J. (a) λ₀ = hc/W₀ = (6.626 × 10^⁻³⁴ × 3.0 × 10^⁸) ÷ (3.044 × 10^⁻¹⁹) ≈ 653 nm. (b) ν₀ = W₀/h ≈ 4.59 × 10^¹⁴ Hz. At 500 nm the photon energy is (6.626 × 10^⁻³⁴ × 3.0 × 10^⁸) ÷ (500 × 10^⁻⁹) = 3.976 × 10^⁻¹⁹ J, so KE = 3.976 × 10^⁻¹⁹ − 3.044 × 10^⁻¹⁹ = 9.32 × 10^⁻²⁰ J (0.58 eV), and v² = 2 KE/m_e = (1.864 × 10^⁻¹⁹ J) ÷ (9.11 × 10^⁻³¹ kg) = 2.05 × 10^¹¹ m^² s^⁻², so v ≈ 4.5 × 10^⁵ m s^⁻¹.

9. Q9. Electrons ejected from silver by radiation of 256.7 nm are just stopped by a reverse voltage of 0.35 V. Find the work function of silver (NCERT Exercise 2.53, page 72).

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   Photon energy = (6.626 × 10^⁻³⁴ × 3.0 × 10^⁸) ÷ (256.7 × 10^⁻⁹) = 7.74 × 10^⁻¹⁹ J = 4.83 eV. The stopping voltage gives KE_max = 0.35 eV. W₀ = 4.83 − 0.35 ≈ 4.48 eV.

10. Q10. Find the wavenumber of the longest-wavelength line in the Balmer series of hydrogen (NCERT Exercise 2.17, page 70).

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    The longest wavelength is the smallest energy gap, n₂ = 3 to n₁ = 2: ν̄ = 109 677 × (1/4 − 1/9) = 109 677 × 5/36 ≈ 1.523 × 10^⁴ cm^⁻¹ (1.523 × 10^⁶ m^⁻¹), a wavelength of about 656 nm.