Question

The dimensional formula of Planck’s constant is:
$\begin{array}{rl}& \text{A}\text{. }\left[M{L}^2{T}^{-1}\right]\\ & \text{B}\text{. }\left[M{L}^2{T}^{-2}\right]\\ & \text{C}\text{. }\left[M{L}^0{T}^2\right]\\ & \text{D}\text{. }\left[ML{T}^{-2}\right]\end{array}$

Answer 1

Hint: We can find the dimensional formula of any constant from a known universal equation. The equation should not have other unknown constants. To derive the dimensional formula of Planck’s constant we can use the energy equation for photon’s energy.

Formula Used:
Energy of photon is given by,
$E=h\nu$
Where,
$h$ is Planck’s constant
$\nu$ is the frequency of light.

Complete answer:
Planck’s constant is one of the most fundamental constants in the domain of quantum mechanics.
So, the dimensional analysis of the Planck’s constant is of utmost interest.
We know that the energy of a photon is given by,
$E=h\nu$
$\Rightarrow h={\displaystyle \frac{E}{\nu }}$
Where,
$h$ is Planck’s constant
$\nu$ is the frequency of light
So, we can write,
$\left[h\right]={\displaystyle \frac{\left[E\right]}{\left[\nu \right]}}$.......................(1)
If we can find the dimensions of energy (E) and frequency (v), then we can find the dimension of Planck’s constant.
Dimensional analysis of basic quantities can be found using the definition.
Energy is given by,
$E=Fs$
Where,
F is the force on an object
s is the distance travelled by an object due to the force
So, we can write,
$\left[E\right]=\left[F\right]\left[s\right]$........................(2)
Force is also given by,
$F=ma$
Hence, the dimension of force is,
$\left[F\right]=\left[m\right]\left[a\right]$
$\Rightarrow \left[F\right]=ML{T}^{-2}$
So, we can put the dimension of force in equation (2).
We have,
$\left[E\right]=ML{T}^{-2}\times L$
$\Rightarrow \left[E\right]=M{L}^2{T}^{-2}$
Now, let's find the dimension of frequency.
Frequency of a wave is given by the inverse of the time period.
Hence, we can write,
$\nu ={\displaystyle \frac{1}{T}}$
So, the dimension of frequency is,
$\left[\nu \right]=T^{-1}$
We can put the dimensions of energy and frequency in equation (1).
So, we have,
$\left[h\right]={\displaystyle \frac{M{L}^2{T}^{-2}}{T^{-1}}}=M{L}^2{T}^{-1}$

Hence, the dimension of Planck’s constant is:
$\left[h\right]=\left[M{L}^2{T}^{-1}\right]$

The correct option is: (A).

Note:
Dimensional analysis of a quantity can be done using any equation which does not contain more than one number of unknown quantities. For example, if there was an equation like the following:
$F=(ma)(kb)$
Where, you know the dimensions of only m and k, then we could not find the dimensions of a or b using this equation.