Question

The de Broglie wavelength of an oxygen molecule at 27°C is x × $10^{-12$ m. The value of x is (take Planck's constant=6.63 × $10^{-34}$ J/s, Boltzmann constant=1.38 × $10^{-23}$ J/K, mass of oxygen molecule = 5.31 × $10^{-26}$ kg)}

• A. 24
• B. 20
• C. 26
• D. 30

Hint:

When calculating square roots of powers of 10, ensure the exponent is even ($10^{-48}$) for easier manual calculation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Thermal de Broglie wavelength is associated with particles in thermal equilibrium. The average kinetic energy of a molecule is related to temperature by $K.E. = \frac{3}{2}kT$.
Step 2: Key Formula or Approach:
1. de Broglie wavelength: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(K.E.)}}$.
2. Thermal K.E.: $\lambda = \frac{h}{\sqrt{3mkT}}$.
Step 3: Detailed Explanation:
Given $T = 27^\circ\text{C} = 300$ K.
$h = 6.63 \times 10^{-34}$.
$m = 5.31 \times 10^{-26}$ kg.
$k = 1.38 \times 10^{-23}$ J/K.
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 5.31 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300}} \] Calculating the denominator: \[ \sqrt{3 \times 5.31 \times 1.38 \times 3 \times 10^{-47}} \approx \sqrt{66 \times 10^{-47}} \approx 25.6 \times 10^{-24} \] \[ \lambda \approx \frac{6.63 \times 10^{-34}}{25.6 \times 10^{-24}} \approx 0.259 \times 10^{-10} = 25.9 \times 10^{-12} \text{ m} \] Comparing with $x \times 10^{-12}$, $x \approx 26$.
Step 4: Final Answer:
The value of x is 26.