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

To find the de-Broglie wavelength of an oxygen molecule at \(27^\circ\)C, we use the de-Broglie wavelength formula:

\[\lambda = \frac{h}{p}\]

where \(h\) is Planck's constant (6.626 × 10^-34 Js) and \(p\) is the momentum of the particle. The momentum can be found using:

\(p = \sqrt{2mK}\]\)

where \(m\) is the mass of the particle, and \(K\) is its kinetic energy. For a molecule in a gas, kinetic energy K at a temperature T is given by:

\(K = \frac{3}{2}k_BT\)

where \(k_B\) is Boltzmann's constant (1.38 × 10^-23 J/K) and \(T\) is the absolute temperature in Kelvin.

1. First, convert the temperature from Celsius to Kelvin:

\(T = 27 + 273 = 300\, \text{K}\)

1. Calculate the mass of one oxygen molecule:
   • The molar mass of oxygen = 32 g/mole = 32 × 10^-3 kg/mole.
   • The mass of one molecule, \(m\), using Avogadro's number \(N_A = 6.022 \times 10^{23}\) molecules/mole:\)
2. Calculate the kinetic energy \(K\):

\(K = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \text{ J}\)

1. Calculate the momentum \(p\):

\(p = \sqrt{2 \times 5.32 \times 10^{-26} \times 6.21 \times 10^{-21}} = 1.61 \times 10^{-23} \text{ kg m/s}\)

1. Finally, calculate the de-Broglie wavelength:

\(\lambda = \frac{6.626 \times 10^{-34}}{1.61 \times 10^{-23}} = 4.11 \times 10^{-11} \text{ m} = 0.257 \text{ \AA}\)

Thus, the de-Broglie wavelength of an oxygen molecule at \(27^\circ\)C is \(0.257\ \text{\AA}\).