Arrange the de-Broglie wavelengths associated with the following in the decreasing order: (Given: $h = 6.6 × 10^{-34} Js$)
(A) an electron (mass $9 × 10^{-31} kg$) in the hydrogen atom moving with the speed of $3 × 10^6 m s^{-1}$
(B) a bullet (mass 0.04 kg) travelling at the speed of $10^3 m s^{-1}$
(C) a ball (mass 0.06 kg) moving with the speed of $1.0 m s^{-1}$
(D) a dust particle (mass $1.0 × 10^{-9} kg$) drifting with the speed of $3.3 m s^{-1}$
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (D), (C), (B)
Formula: $\lambda = \frac{h}{mv}$
Given $h = 6.6 \times 10^{-34} \, Js$
(A) Electron:
$m = 9 \times 10^{-31} \, kg, \; v = 3 \times 10^{6} \, m/s$
$\lambda_A = \frac{6.6 \times 10^{-34}}{(9 \times 10^{-31})(3 \times 10^{6})}$
$= \frac{6.6 \times 10^{-34}}{2.7 \times 10^{-24}} = 2.44 \times 10^{-10} \, m$
(B) Bullet:
$m = 0.04 \, kg, \; v = 10^{3} \, m/s$
$\lambda_B = \frac{6.6 \times 10^{-34}}{0.04 \times 10^{3}}$
$= \frac{6.6 \times 10^{-34}}{40} = 1.65 \times 10^{-35} \, m$
(C) Ball:
$m = 0.06 \, kg, \; v = 1 \, m/s$
$\lambda_C = \frac{6.6 \times 10^{-34}}{0.06}$
$= 1.1 \times 10^{-32} \, m$
(D) Dust particle:
$m = 1.0 \times 10^{-9} \, kg, \; v = 3.3 \, m/s$
$\lambda_D = \frac{6.6 \times 10^{-34}}{3.3 \times 10^{-9}}$
$= 2.0 \times 10^{-25} \, m$
Comparison:
$\lambda_A = 2.44 \times 10^{-10}$
$\lambda_D = 2.0 \times 10^{-25}$
$\lambda_C = 1.1 \times 10^{-32}$
$\lambda_B = 1.65 \times 10^{-35}$
Order (decreasing): A > D > C > B