Question

If a particle of mass \( 6 \times 10^{-30} \) kg is moving with a speed of \( 5.5 \times 10^5 \text{ m s}^{-1} \), then the de Broglie wavelength of the particle is
(Planck's constant = \( 6.6 \times 10^{-34} \text{ J s} \))

• A. 200 \(\text{\AA}\)
• B. 0.2 \(\text{\AA}\)
• C. 2 \(\text{\AA}\)
• D. 20 \(\text{\AA}\)

Hint:

The de Broglie wavelength equation, $\lambda = h/p = h/(mv)$, is fundamental in quantum mechanics, demonstrating the wave-particle duality. Always ensure all values are in consistent SI units before calculation to avoid errors, and pay attention to scientific notation. Conversion to angstroms ($1 \text{ \AA} = 10^{-10} \text{ m}$) is often required for atomic-scale wavelengths.

Answer 1

The Correct Option is C

Step 1: Identify the Formula for de Broglie Wavelength
The de Broglie wavelength ($\lambda$) of a particle is given by the formula: \[ \lambda = \frac{h}{p} \] Where $h$ is Planck's constant and $p$ is the momentum of the particle. The momentum ($p$) of a particle with mass ($m$) and speed ($v$) is given by $p = mv$. Substituting this into the de Broglie wavelength formula: \[ \lambda = \frac{h}{mv} \] Step 2: Extract Given Values and Ensure SI Units
Given: \begin{itemize} \item Mass of the particle, $m = 6 \times 10^{-30} \text{ kg}$ \item Speed of the particle, $v = 5.5 \times 10^5 \text{ m s}^{-1}$ \item Planck's constant, $h = 6.6 \times 10^{-34} \text{ J s}$ \end{itemize} All given values are already in their standard SI units (kg, m/s, J s), so no unit conversion is required at this stage. Step 3: Compute the de Broglie Wavelength (\(\lambda\))
Substitute the given values into the de Broglie wavelength formula: \[ \lambda = \frac{6.6 \times 10^{-34} \text{ J s}}{(6 \times 10^{-30} \text{ kg}) \times (5.5 \times 10^5 \text{ m s}^{-1})} \] First, calculate the denominator (momentum): \[ mv = (6 \times 10^{-30}) \times (5.5 \times 10^5) \] \[ mv = (6 \times 5.5) \times (10^{-30} \times 10^5) \] \[ mv = 33 \times 10^{-25} \text{ kg m s}^{-1} \] Now, calculate the wavelength: \[ \lambda = \frac{6.6 \times 10^{-34}}{33 \times 10^{-25}} \] \[ \lambda = \frac{6.6}{33} \times \frac{10^{-34}}{10^{-25}} \] \[ \lambda = 0.2 \times 10^{-34 - (-25)} \] \[ \lambda = 0.2 \times 10^{-9} \text{ m} \] To express this in angstroms (\(\text{\AA}\)), recall that $1 \text{ \AA} = 10^{-10} \text{ m}$. \[ \lambda = 0.2 \times 10^{-9} \text{ m} = (0.2 \times 10) \times 10^{-10} \text{ m} \] \[ \lambda = 2 \times 10^{-10} \text{ m} \] \[ \lambda = 2 \text{ \AA} \] Step 4: Analyze Options
\begin{itemize} \item Option (1): 200 \(\text{\AA}\). Incorrect. \item Option (2): 0.2 \(\text{\AA}\). Incorrect. \item Option (3): 2 \(\text{\AA}\). Correct, as it matches our calculated wavelength. \item Option (4): 20 \(\text{\AA}\). Incorrect. \end{itemize}