Question

The de Broglie wavelength of a particle is 1000 nm. What is its momentum? ($h=6.6 \times 10^{-34} \text{ J s}$)

• A. $6.6 \times 10^{-25} \text{ J s m}^{-1}$
• B. $6.6 \times 10^{-25} \text{ J s m}$
• C. $6.6 \times 10^{-28} \text{ J s m}^{-1}$
• D. $6.6 \times 10^{-26} \text{ J s m}$

Hint:

de Broglie Relation: Matter exhibits wave-like behavior, and the de Broglie wavelength $\lambda$ is inversely proportional to the momentum $p$: $\lambda = \frachp$.

Answer 1

The Correct Option is C

The de Broglie wavelength of a particle is related to its momentum $p$ by: \[ \lambda = \frac{h}{p} \] Rearranging for $p$: \[ p = \frac{h}{\lambda} \] Substitute the given values:

• $h = 6.6 \times 10^{-34} \text{ J s}$
• $\lambda = 1000 \text{ nm} = 1000 \times 10^{-9} \text{ m} = 10^{-6} \text{ m}$

\[ p = \frac{6.6 \times 10^{-34}}{1 \times 10^{-6}} = 6.6 \times 10^{-28} \text{ J s m}^{-1} \] The units are valid: $\text{J s m}^{-1} = \text{kg m s}^{-1}$, which is correct for momentum. Hence, the correct answer is option (3).

Answer 2

Step 1: Write down the de Broglie wavelength formula
The de Broglie wavelength (λ) of a particle is related to its momentum (p) by the equation:
λ = h / p

Step 2: Rearrange the formula to solve for momentum
Rearranging, we get:
p = h / λ

Step 3: Substitute the given values
Planck's constant, h = 6.6 × 10⁻³⁴ J·s
Wavelength, λ = 1000 nm = 1000 × 10⁻⁹ m = 1 × 10⁻⁶ m

Step 4: Calculate the momentum
p = (6.6 × 10⁻³⁴) / (1 × 10⁻⁶) = 6.6 × 10⁻²⁸ kg·m/s

Step 5: Check units for momentum
J·s·m⁻¹ = (kg·m²/s²)·s·m⁻¹ = kg·m/s, which are the correct units for momentum.

Step 6: Conclusion
The momentum of the particle is 6.6 × 10⁻²⁸ kg·m/s, confirming that the calculation and units are correct.