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

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Modern Physics

The de Broglie wavelength of an oxygen molecule at

2 7^{\circ} C

x \times 1 0^{- 12} m

. The value of x is (take Planck's constant

= 6.63 \times 1 0^{- 34} J\cdots

, Boltzmann constant

= 1.38 \times 1 0^{- 23} J/K

, mass of oxygen molecule

= 5.31 \times 1 0^{- 26} kg

(A)26✓ Correct
(B)24
(C)30
(D)20

Answer: (A)

Solution

At temperature

T

the mean translational kinetic energy is

\frac{3}{2} k T

, so the momentum is

p = 2 m \cdot \frac{3}{2} k T = 3 mk T

and the de Broglie wavelength is

λ = \frac{h}{3 mk T}

. With

T = 300

λ = \frac{6.63 \times 1 0 ^{- 34}}{3 \times 5.31 \times 1 0 ^{- 26} \times 1.38 \times 1 0 ^{- 23} \times 300} = \frac{6.63 \times 1 0 ^{- 34}}{2.57 \times 1 0 ^{- 23}} \approx 2.58 \times 1 0^{- 11}

= 25.8 \times 1 0^{- 12}

m. Hence

x \approx 26

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option

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