Consider an atom traveling at 1% of the speed of light. The de Broglie wavelength of the atom is found to be 3.32 × 10⁻³ pm. Which element is this?

• A. Carbon
• B. Hydrogen
• C. Uranium
• D. Helium

Answer: (C)

Verified step by step guidance

1. First, understand the de Broglie wavelength formula: \( \lambda = \frac{h}{mv} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ m}^2 \text{ kg/s})\), \( m \) is the mass of the particle, and \( v \) is the velocity of the particle.
2. Convert the given de Broglie wavelength from picometers to meters: \( 3.32 \times 10^{-3} \text{ pm} = 3.32 \times 10^{-15} \text{ m} \).
3. Calculate the velocity of the atom, which is 1% of the speed of light. The speed of light \( c \) is approximately \( 3.00 \times 10^8 \text{ m/s} \), so \( v = 0.01 \times c = 3.00 \times 10^6 \text{ m/s} \).
4. Rearrange the de Broglie equation to solve for mass: \( m = \frac{h}{\lambda v} \). Substitute the known values: \( m = \frac{6.626 \times 10^{-34} \text{ m}^2 \text{ kg/s}}{3.32 \times 10^{-15} \text{ m} \times 3.00 \times 10^6 \text{ m/s}} \).
5. Compare the calculated mass with the atomic masses of the given elements. The mass should be closest to the atomic mass of Uranium, which is approximately \( 238 \text{ u} \) (where \( 1 \text{ u} = 1.660539 \times 10^{-27} \text{ kg} \)).