Question

Find de-Broglie wavelength of an oxygen molecule at $27^\circ$C. Molar mass of oxygen molecule is $32$ g/mole.

• A. $0.257\ \text{\AA}$
• B. $2.57\ \text{\AA}$
• C. $25.7\ \text{\AA}$
• D. $257\ \text{\AA}$

Hint:

For thermal motion of gas molecules, directly use $\lambda=\dfrac{h}{\sqrt{3mkT}}$ instead of finding velocity separately.

Answer 1

The Correct Option is A

Concept: The de-Broglie wavelength of a particle is given by: \[ \lambda = \frac{h}{p} \] For gas molecules at temperature $T$, the rms momentum is: \[ p = \sqrt{3mkT} \] Hence, \[ \lambda = \frac{h}{\sqrt{3mkT}} \]
Step 1: Convert given quantities Temperature: \[ T = 27^\circ\text{C} = 300\ \text{K} \] Mass of one oxygen molecule: \[ m = \frac{32\times10^{-3}}{6.02\times10^{23}} \approx 5.31\times10^{-26}\ \text{kg} \]
Step 2: Substitute values \[ \lambda = \frac{6.63\times10^{-34}} {\sqrt{3\times 5.31\times10^{-26}\times 1.38\times10^{-23}\times 300}} \]
Step 3: Calculate \[ \lambda \approx 2.57\times10^{-11}\ \text{m} \] Step 4: Convert to angstrom \[ 1\ \text{\AA} = 10^{-10}\ \text{m} \] \[ \lambda = 0.257\ \text{\AA} \]
Step 5: Hence, the de-Broglie wavelength of the oxygen molecule is: \[ \boxed{0.257\ \text{\AA}} \]