More practice questions

4. Q4. A neutron diffraction experiment uses neutrons of wavelength 800 pm. Find the speed of the neutrons (mass 1.675 × 10^⁻²⁷ kg; NCERT Exercise 2.58, page 72).

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   mλ = 1.675 × 10^⁻²⁷ kg × 800 × 10^⁻¹² m = 1.34 × 10^⁻³⁶ kg m, so v = h/(mλ) = (6.626 × 10^⁻³⁴ J s) ÷ (1.34 × 10^⁻³⁶ kg m) = 494 m s^⁻¹.

5. Q5. A hockey ball of mass 0.1 kg moves at 4.37 × 10^⁵ m s^⁻¹, the speed of a proton accelerated through 1000 V. Find its wavelength (NCERT Exercise 2.60, page 72).

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   λ = (6.626 × 10^⁻³⁴) ÷ (0.1 × 4.37 × 10^⁵) = 1.52 × 10^⁻³⁸ m, far too small to detect.

6. Q6. Show that the circumference of a Bohr orbit of hydrogen is a whole number of de Broglie wavelengths of the electron (NCERT Exercise 2.32, page 71).

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   Bohr’s condition is m_evr = nh/2π, so 2πr = nh/m_ev. The de Broglie wavelength is λ = h/m_ev, so 2πr = nλ: the circumference holds exactly n wavelengths.

7. Q7. An electron and a proton have the same kinetic energy. Which has the longer de Broglie wavelength, and by what factor? (The proton is about 1836 times as heavy.)

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   λ = h/√(2mK), so at the same K, λ ∝ 1/√m. The electron has the longer wavelength, by √1836 ≈ 43 times.

8. Q8. Find the de Broglie wavelength of an electron accelerated from rest through 100 V.

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   2m_eeV = 2 × 9.11 × 10^⁻³¹ kg × 1.602 × 10^⁻¹⁹ C × 100 V = 2.919 × 10^⁻⁴⁷ kg^² m^² s^⁻², whose square root is 5.403 × 10^⁻²⁴ kg m s^⁻¹, so λ = h/√(2m_eeV) = (6.626 × 10^⁻³⁴ J s) ÷ (5.403 × 10^⁻²⁴ kg m s^⁻¹) = 1.23 × 10^⁻¹⁰ m, about 123 pm (or 1.23/√100 nm). This is similar to the spacing of atoms in a crystal, which is why such electrons are diffracted by crystals.

9. Q9. The numerical values of the position and momentum standard deviations are equal when expressed in metres and kg m s^⁻¹. Assuming the uncertainty product takes its minimum value, find both and the minimum velocity spread for mass m.

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   The quantities have different dimensions, so only their numerical values are equated. At the minimum product, write Δx = a m and Δp = a kg m s^⁻¹; solving gives Δp = √(h/4π) = √(6.626 × 10^⁻³⁴ ÷ 12.57) = 7.26 × 10^⁻¹⁸ kg m s^⁻¹ (and Δx = 7.26 × 10^⁻¹⁸ m). Then Δv = Δp/m = (1/2m)√(h/π).

10. Q10. A proton and an α-particle are accelerated from rest through the same potential difference. Find the ratio of their de Broglie wavelengths (take the α-particle’s mass as 4 times and its charge as 2 times the proton’s).

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    λ = h/√(2mqV), so λ_p/λ_α = √(m_αq_α/m_pq_p) = √(4 × 2) = √8 ≈ 2.83. The proton has the longer wavelength.