Question 56 of 62intermediate🔧 ApplyNumerical2 marks

If the velocity of the electron in Bohr’s first orbit is 2.19×102.19 \times 10⁶ ms⁻¹, calculate the de Broglie wavelength associated with it.

Correct Answer

The de Broglie wavelength associated with the electron is 3.32×10−103.32 \times 10^{-10} m (or 332 pm).

Exercise: EXERCISES | Q: 2.59 | (Page 41)
For More Understanding

Explanation

To find the de Broglie wavelength of an electron moving in Bohr's first orbit, we use the de Broglie equation, λ=hmv\lambda = \frac{h}{mv}. We are given the velocity of the electron and can use the standard values for Planck's constant and the mass of an electron as provided in the textbook context.

Solution Steps

  1. Step 1: Write down the given values and constants: Velocity of electron, v=2.19×106 ms−1v = 2.19 \times 10^6 \text{ ms}^{-1} Mass of electron, m=9.1×10−31 kgm = 9.1 \times 10^{-31} \text{ kg} Planck's constant, h=6.626×10−34 Jsh = 6.626 \times 10^{-34} \text{ Js}

  2. Step 2: Use the de Broglie equation to find the wavelength:

    λ=hmv\lambda = \frac{h}{mv}

  3. Step 3: Substitute the values into the equation:

    λ=6.626×10−34 Js(9.1×10−31 kg)×(2.19×106 ms−1)\lambda = \frac{6.626 \times 10^{-34} \text{ Js}}{(9.1 \times 10^{-31} \text{ kg}) \times (2.19 \times 10^6 \text{ ms}^{-1})}

  4. Step 4: Calculate the denominator:

    m×v=19.929×10−25 kg ms−1m \times v = 19.929 \times 10^{-25} \text{ kg ms}^{-1}

  5. Step 5: Divide to find the wavelength:

    λ=6.626×10−3419.929×10−25 m=0.3325×10−9 m=3.32×10−10 m\lambda = \frac{6.626 \times 10^{-34}}{19.929 \times 10^{-25}} \text{ m} = 0.3325 \times 10^{-9} \text{ m} = 3.32 \times 10^{-10} \text{ m}
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