91Ó°ÊÓ

The given data can be listed below as,

• The mass of a bullet is, $m=40\text{g}$.
• The speed of the bullet is, $v=1000\text{m/s}$.

A light behaving as a wave also behave as a particle under appropriate condition, while a particle can behave as a wave. The de Broglie wavelength equation gave the related wave nature of the particle.

The expression to calculate the de-Broglie wavelength of the bullet at that speed is expressed as,

$\mathrm{λ}=\frac{h}{mv}$

Here, $\mathrm{λ}$ is the de-Broglie wavelength of the bullet at that speed and h is the Plank’s constant whose value is $6.63×{10}^{-36}\text{J}·\text{s}$.

Substitute all the known values in the above equation.

$\begin{array}{rcl}\mathrm{λ}& =& \left[\frac{\left(6.63×{10}^{-34}\text{J}·\text{s}\right)}{\left(40\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(1000\text{m/s}\right)}\right]\\ & ≈& 1.66×{10}^{-35}\text{J}·{\text{s}}^{\text{2}}\text{/kg}·\text{m}\\ & ≈& \left(1.66×{10}^{-35}\text{J}·{\text{s}}^{\text{2}}\text{/kg}·\text{m}\right)\left(\frac{1\text{m}}{1\text{J}·{\text{s}}^{\text{2}}\text{/kg}·\text{m}}\right)\\ & ≈& 1.66×{10}^{-35}\text{m}\end{array}$

Thus, the de-Broglie wavelength of the bullet at that speed is $1.66×{10}^{-35}\text{m}$.