The ratio of the dimensions of Planck's constant and that of moment of inertia has the dimensions of

Options

A. angular momentum
B. time
C. velocity
D. frequency

Answer

D

Explanation

We know that, energy of an emitted particle,
`" "E=hvrArrh=(E)/(nu)`
Planck's constant
`h=(["ML"^(2)"T"^(-2)])/("T"^(-1))=["ML"^(2)"T"^(-1)]" "…(i) `
and moment of inertia,
`I=mr^(2)rArrI=["ML"^(2)]" "...(ii)`
On dividing Eq. (i) by Eq. (ii), we get
`(h)/(I)=[("ML"^(2)"T"^(-1))/("ML"^(2))]=["T"^(-1)]=(1)/("T")`
i.e., `" "(h)/(I)=["T"^(-1)]`= frequency of a particle.