Explanation

The de Broglie wavelength of a particle is given by

$$ \lambda = \frac{h}{p} $$

For a gas molecule at temperature $T$, the average kinetic energy is

$$ \frac{3}{2}kT = \frac{1}{2}mv^2 $$

From this,

$$ v = \sqrt{\frac{3kT}{m}} $$

Momentum of the molecule is

$$ p = mv = m\sqrt{\frac{3kT}{m}} = \sqrt{3mkT} $$

Hence the de Broglie wavelength becomes

$$ \lambda = \frac{h}{\sqrt{3mkT}} $$

Given values:

$$ h = 6.63 \times 10^{-34}\,\text{J·s} $$ $$ k = 1.38 \times 10^{-23}\,\text{J/K} $$ $$ T = 27^\circ\text{C} = 300\,\text{K} $$ $$ m = 5.31 \times 10^{-26}\,\text{kg} $$

Substituting,

$$ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 5.31 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300}} $$

Calculate inside the square root:

$$ 3 \times 5.31 \times 1.38 \times 300 \approx 6590 $$

$$ \sqrt{6590 \times 10^{-49}} \approx 8.12 \times 10^{-24.5} $$

Now,

$$ \lambda \approx \frac{6.63 \times 10^{-34}}{2.55 \times 10^{-23}} $$

$$ \lambda \approx 2.6 \times 10^{-11}\,\text{m} $$

$$ \lambda = 26 \times 10^{-12}\,\text{m} $$

Hence,

$$ x = 26 $$

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