Question 55 of 62intermediate🔧 ApplyNumerical2 marks

Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.

Correct Answer

The characteristic velocity associated with the neutron is approximately 4.94×102 ms−14.94 \times 10^2 \text{ ms}^{-1}.

Exercise: EXERCISES | Q: 2.58 | (Page 41)
For More Understanding

Explanation

To find the velocity of the neutron, we use the de Broglie equation, λ=hmv\lambda = \frac{h}{mv}, which relates the wavelength of a particle to its momentum. By rearranging this formula, we can solve for the velocity vv using the given wavelength and the known mass of a neutron.

Solution Steps

  1. Given: Wavelength, λ=800 pm=800×10−12 m=8×10−10 m\lambda = 800 \text{ pm} = 800 \times 10^{-12} \text{ m} = 8 \times 10^{-10} \text{ m} Planck's constant, h=6.626×10−34 Jsh = 6.626 \times 10^{-34} \text{ Js} Mass of a neutron, m=1.675×10−27 kgm = 1.675 \times 10^{-27} \text{ kg}

  2. Formula: According to de Broglie's equation, λ=hmv\lambda = \frac{h}{mv} Rearranging for velocity (vv), we get v=hmλv = \frac{h}{m \lambda}

  3. Calculation:

    v=6.626×10−34 Js(1.675×10−27 kg)×(8×10−10 m)v = \frac{6.626 \times 10^{-34} \text{ Js}}{(1.675 \times 10^{-27} \text{ kg}) \times (8 \times 10^{-10} \text{ m})} v=6.626×10−3413.4×10−37 ms−1v = \frac{6.626 \times 10^{-34}}{13.4 \times 10^{-37}} \text{ ms}^{-1} v=0.4944×103 ms−1v = 0.4944 \times 10^{3} \text{ ms}^{-1} v=4.94×102 ms−1v = 4.94 \times 10^{2} \text{ ms}^{-1}
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