Question 21 of 62intermediate🔧 ApplyNumerical2 marks

The mass of an electron is 9.1×109.1 \times 10⁻³¹ kg. If its K.E. is 3.0×103.0 \times 10⁻²⁵ J, calculate its wavelength.

Correct Answer

8967×10−10 m8967 \times 10^{-10} \text{ m} or 896.7 nm896.7 \text{ nm}

Exercise: EXERCISES | Q: 2.21 | (Page 41)
For More Understanding

Explanation

The velocity of the electron is first determined from its kinetic energy using the formula K.E.=12mv2K.E. = \frac{1}{2}mv^2. Once the velocity is known, the de Broglie equation λ=hmv\lambda = \frac{h}{mv} is used to calculate the wavelength associated with the moving electron.

Solution Steps

  1. Step 1: Calculate the velocity of the electron using the kinetic energy formula.

    K.E.=12mv2K.E. = \frac{1}{2}mv^2 v=(2×K.E.m)1/2=(2×3.0×10−25 kg m2 s−29.1×10−31 kg)1/2=812 m s−1v = \left( \frac{2 \times K.E.}{m} \right)^{1/2} = \left( \frac{2 \times 3.0 \times 10^{-25} \text{ kg m}^{2} \text{ s}^{-2}}{9.1 \times 10^{-31} \text{ kg}} \right)^{1/2} = 812 \text{ m s}^{-1}

  2. Step 2: Calculate the wavelength using the de Broglie equation.

    λ=hmv=6.626×10−34 Js(9.1×10−31 kg)(812 m s−1)\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34} \text{ Js}}{(9.1 \times 10^{-31} \text{ kg})(812 \text{ m s}^{-1})} =8967×10−10 m=896.7 nm= 8967 \times 10^{-10} \text{ m} = 896.7 \text{ nm}
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