A) Energy
B) Linear momentum
C) Work
D) Angular momentum
Correct Answer: B
Solution :
We define energy of a photon\[E=hf,\]here h is Planck's constant and \[f\]is the frequency \[h=\frac{E}{f}\Rightarrow \frac{[E]}{[f]}=\frac{[M{{L}^{2}}{{T}^{-2}}]}{[{{T}^{-1}}]}=[M{{L}^{2}}{{T}^{-1}}]\] We define angular momentum as \[L=I\omega \] \[\Rightarrow \] \[[L]=[I][\omega ]=[M{{L}^{2}}][{{T}^{-1}}]=[M{{L}^{2}}{{T}^{-1}}]\] Hence, \[[h]=[L]=[M{{L}^{2}}{{T}^{-1}}]\]
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