Q56 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
The de Broglie wavelength of an oxygen molecule at $27^{\circ} \mathrm{C}$ is $x \times 10^{-12} \mathrm{~m}$. The value of $x$ is (take Planck's constant $=6.63 \times 10^{-34} \mathrm{~J}. \mathrm{s}$, Boltzmann constant $=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$, mass of oxygen molecule $=5.31 \times 10^{-26} \mathrm{~kg}$)
- A24
- Correct answerB26
- C30
- D20
Solution
The de Broglie wavelength is $\lambda = \frac{h}{mv}$. For an ideal gas, $v_{rms} = \sqrt{\frac{3kT}{m}}$
Therefore: $\lambda = \frac{h}{\sqrt{3mkT}}$
Substituting values: $\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 5.31 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{6.62 \times 10^{-46}}} = \frac{6.63 \times 10^{-34}}{2.57 \times 10^{-23}}$
$\lambda = 2.58 \times 10^{-11}$ m $= 25.8 \times 10^{-12}$ m, so $x \approx 26$
Therefore: $\lambda = \frac{h}{\sqrt{3mkT}}$
Substituting values: $\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 5.31 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{6.62 \times 10^{-46}}} = \frac{6.63 \times 10^{-34}}{2.57 \times 10^{-23}}$
$\lambda = 2.58 \times 10^{-11}$ m $= 25.8 \times 10^{-12}$ m, so $x \approx 26$
Hence, the correct answer is option: 'B'.