Question

“Angular momentum (L) and Planck's constant (h) (where h equals (energy /
frequency)) can be dimensionally equated”. Establish the velocity, or otherwise, of
this statement using dimensional analysis.​

Answer 1

The statement that  “Angular momentum (L) and Planck's constant (h) can be dimensionally equated” is true

Explanation:

We know that

$E=h\nu$

Where E is energy, h is Planck's constant and $\nu$ is frequency

We also know that the dimensions of energy = $[ML^2T^{-2}]$

And the dimensions of frequency is $[T^{-1}]$

Thus, the dimensions of Planck's constant

= Dimensions of Energy/Dimensions of frequency

$=\frac{[ML^2T^{-2}]}{[T^{-1}]}$

$=[ML^2T^{-1}]$

also we know that

Angular momentum

$L=I\omega$

Moment of inertia I is of the form kMR², where k is a constant

Thus, the dimensions of moment of inertia = $[ML^2]$

And the dimensions of angular velocity = $[T^{-1}]$

Therefore, the dimensions of angular momentum

= dimensions of moment of inertia × dimensions of angular velocity

= $[ML^2]\times [T^{-1}]$

= $[ML^2T^{-1}]$

Thus, the statement that  “Angular momentum (L) and Planck's constant (h) can be dimensionally equated” is true.

Hope this answer is helpful.

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