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The dimension of Planck constant equals to that of :
A
Energy
B
Momentum
C
Angular momentum
D
Power
Detailed Solution
From $E = h\nu$: $h = \dfrac{E}{\nu}$
$[h] = \dfrac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1}$
Angular momentum $L = mvr$: $[L] = M \cdot LT^{-1} \cdot L = ML^2T^{-1}$
So Planck's constant has the same dimensions as angular momentum (this is why Bohr's condition $mvr = \dfrac{nh}{2\pi}$ is dimensionally correct).
Energy is $ML^2T^{-2}$, momentum is $MLT^{-1}$ and power is $ML^2T^{-3}$.
$[h] = \dfrac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1}$
Angular momentum $L = mvr$: $[L] = M \cdot LT^{-1} \cdot L = ML^2T^{-1}$
So Planck's constant has the same dimensions as angular momentum (this is why Bohr's condition $mvr = \dfrac{nh}{2\pi}$ is dimensionally correct).
Energy is $ML^2T^{-2}$, momentum is $MLT^{-1}$ and power is $ML^2T^{-3}$.