The dimensional formula of Planck’s constant is

Question

The dimensional formula of Planck’s constant is (A) [.ML- 3T- 1]. (B) [.ML- 2T- 1]. (C) [.M0L- 1T- 3]. (D) [.ML2T- 1].

Tardigrade Admin · Accepted Answer

According to Planck: E = h v h = (E/v) Substituting the dimensions of known physical quantities: [. h ].=([. M L2 T - 2 ]./[. T - 1 ].) or [. h ].=ML2T - 1 So, SI unit or Planckâs constant is kg m2s-1 which can also be written as (. k g m2 s - 2 .) × s . But as k g m2 s - 2 is joule, so unit of h is j o u l e × sec , i.e., Js.

Q. The dimensional formula of Planck’s constant is

$[M L^{- 3} T^{- 1}]$

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

$[M^{0} L^{- 1} T^{- 3}]$

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

Q. The dimensional formula of Planck’s constant is

[ML−3T−1]

[ML−2T−1]

[M0L−1T−3]

[ML2T−1]

According to Planck: E=hv h=vE​ Substituting the dimensions of known physical quantities: [h]=[T−1][ML2T−2]​or[h]=ML2T−1 So, SI unit or Planck’s constant is kg m2s-1 which can also be written as (kgm2s−2)×s . But as kgm2s−2 is joule, so unit of h is joule×sec , i.e., Js.

$[M L^{- 3} T^{- 1}]$

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

$[M^{0} L^{- 1} T^{- 3}]$

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