Question

The de Broglie wavelength of a proton moving at a speed of 1.0 m/s is ______ Å.
[Given: Planck’s constant = 6.626 × 10$^{-34}$ m$^2$kg/s; $m_p$ = 1.67 × 10$^{-27}$ kg]

Hint:

$\lambda = h / (mv)$ — Lighter and slower particles have longer wavelengths.

Answer 1

Correct Answer: 3960

Step 1: Write the de Broglie relation.
\[ \lambda = \frac{h}{mv} \] Step 2: Substitute values.
\[ h = 6.626 \times 10^{-34}~\text{m}^2\text{kg/s}, \quad m = 1.67 \times 10^{-27}~\text{kg}, \quad v = 1.0~\text{m/s} \] \[ \lambda = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 1.0} = 3.97 \times 10^{-7}~\text{m} \] Step 3: Convert meters to Ångström.
\[ 1~\text{Å} = 10^{-10}~\text{m} \Rightarrow 3.97 \times 10^{-7}~\text{m} = 3.97 \times 10^{3}~\text{Å} \] Wait — check unit consistency carefully: \[ 3.97 \times 10^{-7}~\text{m} = 3.97 \times 10^{3}~\text{Å} \] Final Answer: 3.97 × 10$^{3$ Å}

Step 4: Conclusion.
The de Broglie wavelength = 3.97 × 10$^{3$ Å}.