Arrange the following in the increasing order of de-Broglie wavelengt…

The correct answer is Option (4) → A < B < C < E < D

The de Broglie wavelength of a particle depends only on the momentum, so the larger the momentum, the shorter the wavelength.

    \( \displaystyle \lambda = \frac{h}{p}, \qquad h = 6.63\times10^{-34}\,\mathrm{J\,s} \)

A. A bullet of mass 0.02 kg travelling at the speed of 3.3 km/s: \( v = 3.3\,\mathrm{km/s} = 3300\,\mathrm{m/s} \)

    \( \displaystyle p = 0.02 \times 3300 = 66\,\mathrm{kg\,m/s}, \qquad \lambda_{A} = \frac{6.63\times10^{-34}}{66} \approx 1.0\times10^{-35}\,\mathrm{m} \)

B. A ball of mass 0.0331 kg moving with speed of 2 m/s:

    \( \displaystyle p = 0.0331 \times 2 = 0.0662\,\mathrm{kg\,m/s}, \qquad \lambda_{B} = \frac{6.63\times10^{-34}}{0.0662} \approx 1.0\times10^{-32}\,\mathrm{m} \)

C. A dust particle of mass \( 2 \times 10^{-10} \) kg drifting with a speed of 3.3 m/s:

    \( \displaystyle p = 2\times10^{-10} \times 3.3 = 6.6\times10^{-10}\,\mathrm{kg\,m/s}, \qquad \lambda_{C} = \frac{6.63\times10^{-34}}{6.6\times10^{-10}} \approx 1.0\times10^{-24}\,\mathrm{m} \)

D. A photon having a momentum of \( 6.63 \times 10^{-26} \) kg m/s: its momentum is given directly.

    \( \displaystyle \lambda_{D} = \frac{6.63\times10^{-34}}{6.63\times10^{-26}} = 1.0\times10^{-8}\,\mathrm{m} \)

E. An electron accelerated through a potential difference of 100 V: the standard result for an electron gives

    \( \displaystyle \lambda_{E} = \frac{1.227}{\sqrt{V}}\,\mathrm{nm} = \frac{1.227}{\sqrt{100}}\,\mathrm{nm} = 0.1227\,\mathrm{nm} \approx 1.23\times10^{-10}\,\mathrm{m} \)

Putting the five wavelengths in increasing order:

    \( 10^{-35} \lt 10^{-32} \lt 10^{-24} \lt 1.23\times10^{-10} \lt 10^{-8}\ \mathrm{m} \)

The order is A < B < C < E < D.