Dimensions of Planck's Constant (h)
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 (\(f\)).
Dimensions of Energy (E)
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:
- [M] represents mass
- [L] represents length
- [T] represents time
Dimensions of Frequency (f)
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:
\([f] = [T^{-1}]\)
Calculating Dimensions of Planck's Constant (h)
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 \([M L^2 T^{-1}]\).
Analyzing Dimensions of Given Options
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) | \([L] \times [M L T^{-2}] = [M L^2 T^{-2}]\) |
Conclusion
Comparing the dimensions of Planck's constant, \([h] = [M L^2 T^{-1}]\), with the dimensions of the options:
- Linear momentum: \([M L T^{-1}]\) (Does not match)
- Angular momentum: \([M L^2 T^{-1}]\) (Matches)
- Displacement: [L] (Does not match)
- Torque: \([M L^2 T^{-2}]\) (Does not match)
Therefore, the dimensions of Planck's constant (\(h\)) are the same as that of angular momentum.
Similar Questions
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Important Questions from Work, Power & Energy (D)
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