Dual Nature of Matter

JEE Main level, single correct, medium

10 of 200

Question 10 of 200

The de Broglie wavelength of an oxygen molecule at 27∘C27^{\circ} \mathrm{C} is x×10−12 mx \times 10^{-12} \mathrm{~m}. The value of xx is (take Planck's constant =6.63×10−34 J.s=6.63 \times 10^{-34} \mathrm{~J}. \mathrm{s}, Boltzmann constant =1.38×10−23 J/K=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}, mass of oxygen molecule =5.31×10−26 kg=5.31 \times 10^{-26} \mathrm{~kg})

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Answer: (B) 26

The de Broglie wavelength is λ=hmv\lambda = \frac{h}{mv}. For an ideal gas, vrms=3kTmv_{rms} = \sqrt{\frac{3kT}{m}}
Therefore: λ=h3mkT\lambda = \frac{h}{\sqrt{3mkT}}
Substituting values: λ=6.63×10−343×5.31×10−26×1.38×10−23×300\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 5.31 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300}}
λ=6.63×10−346.62×10−46=6.63×10−342.57×10−23\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{6.62 \times 10^{-46}}} = \frac{6.63 \times 10^{-34}}{2.57 \times 10^{-23}}
λ=2.58×10−11\lambda = 2.58 \times 10^{-11} m =25.8×10−12= 25.8 \times 10^{-12} m, so x≈26x \approx 26

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