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MCQEasyJEE 2026de Broglie Relation

JEE Physics 2026 Question with Solution

The de Broglie wavelength of an oxygen molecule at 27∘C27^\circ\text{C} is x×10−12 mx \times 10^{-12} \, \text{m}. The value of xx is (take Planck's constant=6.63×10−34 J/s6.63 \times 10^{-34} \, \text{J/s}, Boltzmann constant=1.38×10−23 J/K1.38 \times 10^{-23} \, \text{J/K}, mass of oxygen molecule = 5.31×10−26 kg5.31 \times 10^{-26} \, \text{kg})

  • A

    2424

  • B

    2020

  • C

    2626

  • D

    3030

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Temperature of the oxygen molecule is 27∘C=300 K27^\circ\text{C} = 300 \, \text{K}. Planck's constant is h=6.63×10−34 J/sh = 6.63 \times 10^{-34} \, \text{J/s}, Boltzmann constant is k=1.38×10−23 J/Kk = 1.38 \times 10^{-23} \, \text{J/K}, and mass of oxygen molecule is m=5.31×10−26 kgm = 5.31 \times 10^{-26} \, \text{kg}.

Find: The value of xx in λ=x×10−12 m\lambda = x \times 10^{-12} \, \text{m}.

For a molecule in thermal equilibrium, the average kinetic energy is

K.E.=32kTK.E. = \frac{3}{2}kT

Using de Broglie relation,

λ=hp=h2m(K.E.)\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(K.E.)}}

Substituting K.E.=32kTK.E. = \frac{3}{2}kT,

λ=h3mkT\lambda = \frac{h}{\sqrt{3mkT}}

Now substitute the given 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}}

Calculating the denominator,

3×5.31×1.38×3×10−47≈66×10−47≈25.6×10−24\sqrt{3 \times 5.31 \times 1.38 \times 3 \times 10^{-47}} \approx \sqrt{66 \times 10^{-47}} \approx 25.6 \times 10^{-24}

Therefore,

λ≈6.63×10−3425.6×10−24≈0.259×10−10=25.9×10−12 m\lambda \approx \frac{6.63 \times 10^{-34}}{25.6 \times 10^{-24}} \approx 0.259 \times 10^{-10} = 25.9 \times 10^{-12} \, \text{m}

Comparing with x×10−12 mx \times 10^{-12} \, \text{m}, we get x≈26x \approx 26.

Therefore, the correct option is C.

Power of Ten Simplification

Given: Use the thermal de Broglie formula

λ=h3mkT\lambda = \frac{h}{\sqrt{3mkT}}

Find: Estimate xx quickly.

First convert 27∘C27^\circ\text{C} to 300 K300 \, \text{K}. Then combine 300=3×102300 = 3 \times 10^2 with the powers of ten:

10−26×10−23×102=10−4710^{-26} \times 10^{-23} \times 10^2 = 10^{-47}

So the denominator has a square root involving approximately

66×10−4766 \times 10^{-47}

Rewrite this in a square-root-friendly way so the power of 1010 is easy to handle. This gives a denominator of about 25.6×10−2425.6 \times 10^{-24}. Then

λ≈6.63×10−3425.6×10−24≈25.9×10−12 m\lambda \approx \frac{6.63 \times 10^{-34}}{25.6 \times 10^{-24}} \approx 25.9 \times 10^{-12} \, \text{m}

Hence x≈26x \approx 26.

This shortcut works because estimating the power of 1010 correctly makes the numerical calculation much faster while preserving the correct order of magnitude. Therefore, the correct option is C.

Common mistakes

  • Using T=27T = 27 instead of converting to 300 K300 \, \text{K} is incorrect because thermal energy formulas require absolute temperature. Always convert degree Celsius to Kelvin before substitution.

  • Using λ=h2mkT\lambda = \frac{h}{\sqrt{2mkT}} directly is incorrect here because the solution uses average thermal kinetic energy K.E.=32kTK.E. = \frac{3}{2}kT. First substitute the thermal kinetic energy, then obtain λ=h3mkT\lambda = \frac{h}{\sqrt{3mkT}}.

  • Handling the power of 1010 incorrectly inside the square root leads to a wrong order of magnitude. Combine exponents carefully before taking the square root, and check that the final wavelength comes in the 10−12 m10^{-12} \, \text{m} range.

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