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NEET CHEMISTRYEasy

A 0.66 kg0.66 \text{ kg} ball is moving with a speed of 100 m/s100 \text{ m/s}. The associated wavelength will be: (h=6.6×10−34 Jsh = 6.6 \times 10^{-34} \text{ Js})

A

6.6×10−34 m6.6 \times 10^{-34} \text{ m}

B

1.0×10−35 m1.0 \times 10^{-35} \text{ m}

C

1.0×10−32 m1.0 \times 10^{-32} \text{ m}

D

6.6×10−32 m6.6 \times 10^{-32} \text{ m}

Step-by-Step Solution

According to the de Broglie equation, the wavelength (λ\lambda) associated with a material particle of mass mm moving with velocity vv is given by: λ=hmv\lambda = \frac{h}{mv}

Given: Mass, m=0.66 kgm = 0.66 \text{ kg} Velocity, v=100 m/sv = 100 \text{ m/s}

  • Planck's constant, h=6.6×10−34 Jsh = 6.6 \times 10^{-34} \text{ Js} (Note: The standard value is 6.626×10−346.626 \times 10^{-34}, but the question specifies 6.66.6).

Calculation: λ=6.6×10−34 kg m2s−10.66 kg×100 m s−1\lambda = \frac{6.6 \times 10^{-34} \text{ kg m}^2\text{s}^{-1}}{0.66 \text{ kg} \times 100 \text{ m s}^{-1}} λ=6.6×10−3466\lambda = \frac{6.6 \times 10^{-34}}{66} λ=0.1×10−34 m\lambda = 0.1 \times 10^{-34} \text{ m} λ=1.0×10−35 m\lambda = 1.0 \times 10^{-35} \text{ m}

This problem illustrates that for macroscopic objects (like a ball), the de Broglie wavelength is extremely small and not observable, consistent with the discussion in the NCERT text .

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