Question

Difficulty: MediumDimensions of Physical Quantities and Dimensional Analysis

In atomic physics, the energy EE of a photon is related to its frequency ff by the formula E=hfE = h f, where hh is Planck's constant. What are the fundamental dimensions of Planck's constant hh?

  1. ML2T−1M L^2 T^{-1}Answer
  2. B
    ML2T−2M L^2 T^{-2}
  3. C
    MLT−1M L T^{-1}
  4. D
    ML2T−3M L^2 T^{-3}

Answer

The fundamental dimensions of Planck's constant are ML2T−1M L^2 T^{-1}.
Planck's constant hh is given by h=E/fh = E / f. Since energy EE has fundamental dimensions of ML2T−2M L^2 T^{-2} and frequency ff has dimensions of T−1T^{-1}, dividing energy by frequency yields ML2T−2T−1=ML2T−1\frac{M L^2 T^{-2}}{T^{-1}} = M L^2 T^{-1}.

Step-by-Step Solution

1
Express Planck's constant in terms of energy and frequency
h=Efh = \frac{E}{f}
Rearranging the equation E=hfE = h f isolates hh.
2
Determine the fundamental dimensions of energy (EE) and frequency (ff)
[E]=ML2T−2[E] = M L^2 T^{-2} and [f]=T−1[f] = T^{-1}
Energy has dimensions of work (F×d=MLT−2⋅L=ML2T−2F \times d = M L T^{-2} \cdot L = M L^2 T^{-2}) and frequency is inverse time (T−1T^{-1}).
3
Divide the dimensions of energy by the dimensions of frequency
[h]=ML2T−2T−1=ML2T−2−(−1)=ML2T−1[h] = \frac{M L^2 T^{-2}}{T^{-1}} = M L^2 T^{-2 - (-1)} = M L^2 T^{-1}
Applying the rules of indices simplifies the exponent of time.

Key Concept

Dimensional Analysis of Physical Constants
Estimated Time:1m 0s
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