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NCERT Exemplar · Q33

Q.According to de Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties. However, a cricket ball of mass 100 g does not move like a wave when it is thrown by a bowler at a speed of 100 km/h. Calculate the wavelength of the ball and explain why it does not show wave nature.

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De Broglie’s wavelength for a macroscopic object is far too small to detect — the cricket ball’s wavelength is about 1.5×10−341.5 \times 10^{-34} m, which is negligible compared to atomic scales, so no wave behaviour is observed.

The key idea is that de Broglie’s relation λ=h/p\lambda = h/p applies to all matter, but the wavelength becomes significant only when the momentum is extremely small — typically for subatomic particles. For a cricket ball, the momentum is enormous on the quantum scale, making the wavelength vanishingly tiny. Let’s work through the numbers.

  1. Convert all quantities to SI units.

    Mass m=100m = 100 g =0.1= 0.1 kg.

    Speed v=100v = 100 km/h. To convert: 11 km/h =10003600= \frac{1000}{3600} m/s =518= \frac{5}{18} m/s.

    So v=100×518=50018≈27.78v = 100 \times \frac{5}{18} = \frac{500}{18} \approx 27.78 m/s.

  2. Find the momentum.

    Momentum p=mv=0.1×27.78=2.778p = mv = 0.1 \times 27.78 = 2.778 kg m/s.

    (We’ll keep it as p=259p = \frac{25}{9} kg m/s for exactness: 0.1×50018=5018=2590.1 \times \frac{500}{18} = \frac{50}{18} = \frac{25}{9}.)

  3. Apply de Broglie’s relation.

    Planck’s constant h=6.626×10−34h = 6.626 \times 10^{-34} J s.

λ=hp=6.626×10−342.778≈2.385×10−34 m.\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{2.778} \approx 2.385 \times 10^{-34} \text{ m}.

Using the exact fraction: λ=6.626×10−3425/9=6.626×10−34×925≈2.385×10−34\lambda = \frac{6.626 \times 10^{-34}}{25/9} = 6.626 \times 10^{-34} \times \frac{9}{25} \approx 2.385 \times 10^{-34} m.

Tip

A quick way to estimate: p≈2.78p \approx 2.78 kg m/s, so λ≈6.6×10−342.8≈2.4×10−34\lambda \approx \frac{6.6 \times 10^{-34}}{2.8} \approx 2.4 \times 10^{-34} m. No calculator needed for the order of magnitude.

  1. Interpret the result. …

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