Question 54 of 62intermediate🔧 ApplyNumerical2 marks

Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6×101.6 \times 10⁶ ms⁻¹, calculate de Broglie wavelength associated with this electron.

Correct Answer

The de Broglie wavelength associated with the electron is 455 pm455 \text{ pm} (or 4.55×10−10 m4.55 \times 10^{-10} \text{ m}).

Exercise: EXERCISES | Q: 2.57 | (Page 41)
For More Understanding

Explanation

According to de Broglie's equation, the wavelength of a particle in motion is inversely proportional to its momentum. By substituting the known values of Planck's constant, the mass of an electron, and the given velocity into the equation, we can calculate the associated wavelength.

Solution Steps

  1. Given values: Mass of electron, m=9.1×10−31 kgm = 9.1 \times 10^{-31} \text{ kg} Velocity of electron, v=1.6×106 ms−1v = 1.6 \times 10^6 \text{ ms}^{-1} Planck's constant, h=6.626×10−34 Jsh = 6.626 \times 10^{-34} \text{ Js}

  2. Formula: According to de Broglie's equation (2.22), λ=hmv\lambda = \frac{h}{mv}

  3. Substitution and Calculation:

    λ=6.626×10−34 Js(9.1×10−31 kg)×(1.6×106 ms−1)\lambda = \frac{6.626 \times 10^{-34} \text{ Js}}{(9.1 \times 10^{-31} \text{ kg}) \times (1.6 \times 10^6 \text{ ms}^{-1})} λ=6.626×10−3414.56×10−25 m\lambda = \frac{6.626 \times 10^{-34}}{14.56 \times 10^{-25}} \text{ m} λ=0.4552×10−9 m\lambda = 0.4552 \times 10^{-9} \text{ m}

  4. Final Answer: λ=4.55×10−10 m\lambda = 4.55 \times 10^{-10} \text{ m} or 455 pm455 \text{ pm}

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