A) frequency
B) velocity
C) angular momentum
D) rime
Correct Answer: A
Solution :
E = hv \[\Rightarrow h=\text{Planck }\!\!'\!\!\text{ s}\,\text{constant}\,\,\text{=}\frac{\text{E}}{\text{v}}\] \[\therefore [h]=\frac{[E]}{[v]}=\frac{[M{{L}^{2}}{{T}^{-2}}]}{[{{T}^{-1}}]}\] \[=[M{{L}^{2}}{{T}^{-1}}]\] and I = moment of inertia = MR2 \[\Rightarrow \,\,\,\,[I]=[M]\,[{{L}^{2}}]\,\,=[M{{L}^{2}}]\] Hence, \[\frac{[h]}{[I]}=\frac{[M{{L}^{2}}{{T}^{-1}}]}{[M{{L}^{2}}]}=[{{T}^{-1}}]\] \[=\frac{1}{[T]}=\dimension\,of\,frequency\] Alternative: \[\frac{h}{I}=\frac{E/v}{I}\] \[=\frac{E\times T}{I}=\frac{(kg\,{{m}^{2}}/{{s}^{2}})\times s}{(kg\,{{m}^{2}})}\] \[=\frac{1}{s}=\frac{1}{time}=\text{frequency}\] Thus, dimensions of \[\frac{h}{I}\] is same of frequency.
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