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

To calculate the de Broglie wavelength of the oxygen molecule at 27°C, we'll use the de Broglie wavelength formula:

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

where:

• \(h = \text{Planck's constant} = 6.63 \times 10^{-34} \, \text{J s}\)
• \(p = \text{momentum of the particle} = mv\)

The velocity \(v\) can be found from the root mean square speed of the gas molecules:

\(v = \sqrt{\frac{3kT}{m}}\)

where:

• \(k = \text{Boltzmann constant} = 1.38 \times 10^{-23} \, \text{J/K}\)
• \(T = \text{Temperature in Kelvin} = (27 + 273) \, \text{K} = 300 \, \text{K}\)
• \(m = \text{mass of oxygen molecule} = 5.31 \times 10^{-26} \, \text{kg}\)

Substitute the values to find \(v\):

\(v = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{5.31 \times 10^{-26}}}\)

Calculating the above expression:

\(v = \sqrt{\frac{1.242 \times 10^{-20}}{5.31 \times 10^{-26}}} = \sqrt{2.34 \times 10^{5}} = 484.35 \, \text{m/s}\)

Now we can calculate the momentum:

\(p = mv = 5.31 \times 10^{-26} \times 484.35 = 2.573 \times 10^{-23} \, \text{kg m/s}\)

Finally, use the de Broglie equation to find \(\lambda\):

\(\lambda = \frac{6.63 \times 10^{-34}}{2.573 \times 10^{-23}} = 2.58 \times 10^{-11} \, \text{m}\)

Convert this wavelength into the given form:

\(2.58 \times 10^{-11} \, \text{m} = 25.8 \times 10^{-12} \, \text{m}\)

Thus, the value of \(x\) is approximately 26. Therefore, the correct answer is 26.