A) frequency
B) velocity
C) angular momentum
D) time
Correct Answer: A
Solution :
| \[E=hv\] |
| \[\Rightarrow \] \[h=\] Plancks constant \[\text{=}\frac{\text{E}}{\text{v}}\] |
| \[\therefore \] \[[h]=\frac{[E]}{[v]}=\frac{[M{{L}^{2}}{{T}^{-2}}]}{[{{T}^{-1}}]}\] |
| \[=[M{{L}^{2}}{{T}^{-1}}]\] |
| and I = moment of inertia \[=M{{R}^{2}}\] |
| \[\Rightarrow \] \[\,[I]=[M]\,[{{L}^{2}}]\,\,=[M{{L}^{2}}]\] |
| Hence, \[\frac{[h]}{[I]}=\frac{[M{{L}^{2}}{{T}^{-1}}]}{[M{{L}^{2}}]}=[{{T}^{-1}}]\] |
| \[=\frac{1}{[T]}=\]dimension of frequency |
| Alternative: \[\frac{h}{I}=\frac{E/v}{I}\] |
| \[=\frac{E\times T}{I}=\frac{(kg\,{{m}^{2}}/{{s}^{2}})\times s}{(kg\,{{m}^{2}})}\] |
| \[=\frac{1}{s}=\frac{1}{\text{time}}=\text{frequency}\] |
| Thus, dimensions of \[\frac{h}{I}\] is same of frequency. |
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