The question asks us to determine the dimensions of Planck's constant, denoted by '\(h\)', given the relationship between the energy (\(E\)) of a photon and its frequency (\(f\)):
E = hf
To find the dimensions of '\(h\)', we can rearrange the formula:
\(h = \frac{E}{f}\)
Now, let's determine the dimensions of energy (\(E\)) and frequency (\(f\)).
Energy is the capacity to do work. Common units of energy include Joules (J). In terms of base SI units, 1 Joule is equal to 1 \(kg \cdot m^2 / s^2\). Therefore, the dimensions of energy are:
\([E] = [M L^2 T^{-2}]\)
Where:
Frequency is the number of cycles per unit time. Its unit is Hertz (Hz), which is equivalent to \(s^{-1}\). Therefore, the dimensions of frequency are:
\([f] = [T^{-1}]\)
Using the formula h = E/f, we can substitute the dimensions we found:
\([h] = \frac{[E]}{[f]} = \frac{[M L^2 T^{-2}]}{[T^{-1}]}\)
Simplifying this expression gives:
\([h] = [M L^2 T^{-2} \cdot T^1] = [M L^2 T^{-1}]\)
So, the dimensions of Planck's constant '\(h\)' are \([M L^2 T^{-1}]\).
Now, let's find the dimensions of each option provided and compare them with the dimensions of '\(h\)'.
| Quantity | Formula/Definition | Dimensions |
|---|---|---|
| Linear Momentum | p = mv (mass × velocity) | \([M] \times [L T^{-1}] = [M L T^{-1}]\) |
| Angular Momentum | L = mvr (mass × velocity × radius) | \([M] \times [L T^{-1}] \times [L] = [M L^2 T^{-1}]\) |
| Displacement | Change in position | [L] |
| Torque | \(\tau = rF\) (radius × Force) | \([L] \times [M L T^{-2}] = [M L^2 T^{-2}]\) |
Comparing the dimensions of Planck's constant, \([h] = [M L^2 T^{-1}]\), with the dimensions of the options:
Therefore, the dimensions of Planck's constant (\(h\)) are the same as that of angular momentum.