The dimensions of Planck-s constant are the same as that of theworklinear impulselinear momentumangular

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Toppr Admin · Accepted Answer

Energy of a photon- E-h-x3C5-where h is the Planck-apos-s constant and -x3C5-xA0-is the frequency-x2234-h-E-v-ML2T-x2212-2-T-x2212-1-ML2T-x2212-1-Angular momentum - Moment of inertia x Angular velocity -Angular momentum-xA0- -ML2-T-x2212-1-ML2T-x2212-1-

The dimensions of Planck's constant are the same as that of the
work
linear impulse
linear momentum
angular

Energy of a photon, $E = h υ$

where h is the Planck's constant and $υ$ is the frequency.

$∴ [h] = \frac{[E]}{[v]} = \frac{[M L^{2} T^{- 2}]}{[T^{- 1}]} = [M L^{2} T^{- 1}]$

Angular momentum = Moment of inertia x Angular velocity (Angular momentum)
= $[M L^{2}] [T^{- 1}] = [M L^{2} T^{- 1}]$

The dimensions of Planck's constant are the same as that of theworklinear impulselinear momentumangular

Energy of a photon, E=hυwhere h is the Planck's constant and υ is the frequency.∴[h]=[E][v]=[ML2T−2][T−1]=[ML2T−1]Angular momentum = Moment of inertia x Angular velocity (Angular momentum) = [ML2][T−1]=[ML2T−1]

The dimensions of Planck's constant are the same as that of the
work
linear impulse
linear momentum
angular

Energy of a photon, $E = h υ$

where h is the Planck's constant and $υ$ is the frequency.

$∴ [h] = \frac{[E]}{[v]} = \frac{[M L^{2} T^{- 2}]}{[T^{- 1}]} = [M L^{2} T^{- 1}]$

Angular momentum = Moment of inertia x Angular velocity (Angular momentum)
= $[M L^{2}] [T^{- 1}] = [M L^{2} T^{- 1}]$