The dimensional formula for Planck's constant | vedclass

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Question

(C) The energy of a photon is given by the equation $E = h 
u$,where $E$ is energy,$h$ is Planck's constant,and $
u$ is frequency.Rearranging for $h

Vedclass · Accepted Answer

$M L^2 T^{-1}$

Vedclass · Answer

(C) The energy of a photon is given by the equation $E = h 
u$,where $E$ is energy,$h$ is Planck's constant,and $
u$ is frequency.Rearranging for $h$,we get $h = \frac{E}{
u}$.The dimensional formula for energy $E$ is $[M L^2 T^{-2}]$.The dimensional formula for frequency $
u$ is $[T^{-1}]$.Substituting these into the equation: $[h] = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}]$.Therefore,the correct option is $C$.

Column-$1$	Column-$2$
$(a)$ Wave vector	$(i)$ $M^1L^0T^{-3}$
$(b)$ Linear mass density	$(ii)$ $M^1L^{-1}T^{-2}$
$(c)$ Intensity of a wave	$(iii)$ $M^0L^{-1}T^0$
$(d)$ Volume elasticity coefficient	$(iv)$ $M^1L^{-1}T^0$

The dimensional formula for Planck's constant $(h)$ is

Similar Questions

Match Column-$1$ with Column-$2$.
Column-$1$ Column-$2$
$(a)$ Wave vector $(i)$ $M^1L^0T^{-3}$
$(b)$ Linear mass density $(ii)$ $M^1L^{-1}T^{-2}$
$(c)$ Intensity of a wave $(iii)$ $M^0L^{-1}T^0$
$(d)$ Volume elasticity coefficient $(iv)$ $M^1L^{-1}T^0$

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