JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

The de Broglie wavelength of an oxygen molecule at $2 7^{\circ} C$ is $x \times 1 0^{- 12} m$ . The value of $x$ is (take Planck's constant $= 6.63 \times 1 0^{- 34} J.s$ , Boltzmann constant $= 1.38 \times 1 0^{- 23} J/K$ , mass of oxygen Molecule $= 5.31 \times 1 0^{- 26} kg$ ).

A
$26$
B
$24$
C
$30$
D
$20$

Solution

Average kinetic energy of a gas molecule at temperature $T$ is $\frac{3}{2} k T$ , so $p = 2 m K = 3 mk T$ , giving:

$λ = \frac{h}{3 mk T}$

With $T = 300 K$ :

$λ = \frac{6.63 \times 1 0 ^{- 34}}{3 \times 5.31 \times 1 0 ^{- 26} \times 1.38 \times 1 0 ^{- 23} \times 300}$

The denominator $= 6.594 \times 1 0^{- 46} \approx 2.568 \times 1 0^{- 23}$ .

$λ \approx 2.58 \times 1 0^{- 11} m = 25.8 \times 1 0^{- 12} m$

$∴ x \approx 26$ .

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →