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MCQEasyJEE 2024Dimensions & Dimensional Analysis

JEE Physics 2024 Question with Solution

The de-Broglie wavelength associated with a particle of mass mm and energy EE is:

λ=h2mE\lambda = \frac{h}{\sqrt{2mE}}

The dimensional formula for Planck’s constant is:

  • A

    [ML−1T−2][ML^{-1}T^{-2}]

  • B

    [ML2T−1][ML^{2}T^{-1}]

  • C

    [MLT−2][MLT^{-2}]

  • D

    [ML2T−2][ML^{2}T^{-2}]

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

  • de-Broglie relation:
λ=h2mE\lambda = \frac{h}{\sqrt{2mE}}
  • λ\lambda is wavelength, mm is mass, and EE is energy.

Find: The dimensional formula of Planck’s constant hh.

Rearrange the given equation:

h=λ2mEh = \lambda \sqrt{2mE}

Now write dimensions of each quantity:

  • [λ]=[L][\lambda] = [L]
  • [m]=[M][m] = [M]
  • [E]=[ML2T−2][E] = [ML^2T^{-2}]

So,

2mE=[M][ML2T−2]=[M2L2T−2]=[MLT−1]\sqrt{2mE} = \sqrt{[M][ML^2T^{-2}]} = \sqrt{[M^2L^2T^{-2}]} = [MLT^{-1}]

Therefore,

[h]=[L]×[MLT−1]=[ML2T−1][h] = [L] \times [MLT^{-1}] = [ML^2T^{-1}]

Hence, the dimensional formula of Planck’s constant is [ML2T−1][ML^2T^{-1}]. The correct option is B.

The solution also states that the correct answer is Option (2), which matches B.

Dimensional Analysis Expansion

Starting from

λ=h2mE\lambda = \frac{h}{\sqrt{2mE}}

we isolate hh as

h=λ2mEh = \lambda \sqrt{2mE}

The constant 22 is dimensionless, so it does not affect dimensions. Therefore,

[h]=[λ]⋅[m][E][h] = [\lambda] \cdot \sqrt{[m][E]}

Substitute the known dimensions:

[h]=[L]⋅[M][ML2T−2][h] = [L] \cdot \sqrt{[M][ML^2T^{-2}]} [h]=[L]⋅[M2L2T−2][h] = [L] \cdot \sqrt{[M^2L^2T^{-2}]}

Taking the square root gives

[M2L2T−2]=[MLT−1]\sqrt{[M^2L^2T^{-2}]} = [MLT^{-1}]

Hence,

[h]=[L][MLT−1]=[ML2T−1][h] = [L][MLT^{-1}] = [ML^2T^{-1}]

Therefore, Planck’s constant has dimensional formula [ML2T−1][ML^2T^{-1}], so the correct option is B.

Common mistakes

  • Using the dimensional formula of momentum instead of Planck’s constant is incorrect. The relation must first be rearranged to solve for hh. Start from h=λ2mEh = \lambda \sqrt{2mE}, not from a guessed standard formula.

  • Forgetting that energy has dimensions [ML2T−2][ML^2T^{-2}] leads to a wrong final power of LL or TT. Always substitute the full energy dimension before simplifying.

  • Treating the square root incorrectly is a common error. Since [M2L2T−2]=[MLT−1]\sqrt{[M^2L^2T^{-2}]} = [MLT^{-1}], each exponent must be halved inside the square root.

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