A) energy
B) power
C) impulse
D) angular momentum
Correct Answer: D
Solution :
From the relation, we observe \[E=hv\] or, \[h=\frac{E}{v}=\frac{M{{L}^{2}}{{T}^{2}}}{1/T}=M{{L}^{2}}{{T}^{-1}}\] ...(1) Again, angular momentum \[L=l\omega \] \[=M{{L}^{2}}\times \frac{1}{T}=M{{L}^{2}}{{T}^{-1}}\] ...(2) Hence, from eqs. (1) and (2) Dimensions of h = dimensions of angular momentum\[=ML{{\,}^{2}}{{T}^{-1}}\] Hence, Plancks constant has the dimensions of angular momentum.
You need to login to perform this action.
You will be redirected in
3 sec
