Question

The de Broglie wavelength of an argon atom (mass = 40 amu) traveling at a speed of 250 m/s (rounded off to one decimal place) is \(\underline{\hspace{2cm}}\) picometers.
Given: \( N = 6.022 \times 10^{23}, h = 6.626 \times 10^{-34} \, \text{kg m}^2 \text{s}^{-1} \).

Hint:

The de Broglie wavelength of a particle can be found using the formula \( \lambda = \frac{h}{mv} \), where \( m \) is the particle mass and \( v \) is its velocity.

Answer 1

Correct Answer: 39.5

The de Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{mv} \] where \( m \) is the mass of the argon atom and \( v \) is its velocity.
The mass of 1 argon atom is:
\[ m = \frac{40}{N} = \frac{40}{6.022 \times 10^{23}} = 6.64 \times 10^{-23} \, \text{kg}. \] Substituting into the de Broglie equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(6.64 \times 10^{-23})(250)} = 4.00 \times 10^{-12} \, \text{m}. \] Converting to picometers:
\[ \lambda = 4.00 \, \text{pm}. \] Thus, the de Broglie wavelength is \( 4.0 \, \text{pm} \).