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Question

(D) The energy of a photon is given by $E = h
u$,where $h$ is Planck's constant and $
u$ is the frequency.Thus,the dimension of $h$ is $[h] = [E] /

Vedclass · Accepted Answer

Angular momentum

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(D) The energy of a photon is given by $E = h
u$,where $h$ is Planck's constant and $
u$ is the frequency.Thus,the dimension of $h$ is $[h] = [E] / [
u]$.$[E] = [M L^2 T^{-2}]$ and $[
u] = [T^{-1}]$.So,$[h] = [M L^2 T^{-2}] / [T^{-1}] = [M L^2 T^{-1}]$.Angular momentum $L$ is defined as $L = mvr$,where $m$ is mass,$v$ is velocity,and $r$ is radius.$[L] = [M] [L T^{-1}] [L] = [M L^2 T^{-1}]$.Therefore,the dimension of Planck's constant is equal to that of angular momentum.

The dimension of Planck's constant is equal to that of:

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