Question

According to Quantum theory of radiation, the energy $$E$$ of a photon of frequency $$\nu$$ is given by $$E=h\nu$$ where $$h$$ is Planck’s constant. The dimensional formula for $$h$$ is:
$$\left[M^1{L}^2{T}^{-2}\right]$$
$$\left[M^1{L}^2{T}^{-1}\right]$$
$$\left[M^1{L}^2{T}^1\right]$$
$$\left[M^1{L}^2{T}^2\right]$$

Answer 1

Hint:
First of all, we will rearrange the given equation and then will separately find the dimensions of energy and frequency. We will substitute the required values in the expression and manipulate accordingly to obtain the result.

Complete step by step solution:
In the given question, we are provided the following data:
$$E=h\nu$$ …… (1)
Where,
$$E$$ indicates energy.
$$h$$ indicates Planck’s constant.
$$\nu$$ indicates frequency.
We are asked to find the dimension of the Planck’s content i.e. we need to find the fundamental quantities associated with the Planck’s constant.
Now, we will find the dimension of the individual physical quantities present in the equation (1):
First, we will go for energy:
Energy is equivalent to work done $$\left(W\right)$$ . So, let us first break down work done into all the associated physical quantities to the ground level.
  $E=W$
  $E=F\times s$
Where,
$$F$$ indicates force in Newtons.
$$s$$ indicates distance.
Again, we break down the above expression:
$$\begin{gathered} E=F\times s \\ E=m\times a\times s \end{gathered}$$
Where,
$$m$$ indicates mass of the body.
$$a$$ indicates acceleration.

Now, we write the dimension of all the physical quantities found in the last equation:
  $E=m\times a\times s$
 $E=\left[M^1{L}^0{T}^0\right]\times \left[M^0{L}^1{T}^{-2}\right]\times \left[M^0{L}^1{T}^0\right]$
$E=\left[M^1{L}^2{T}^{-2}\right]$

Again, we find the dimension of frequency, which is the reciprocal of time.
 $$\begin{gathered} \nu ={\displaystyle \frac{1}{t}} \\ \nu =t^{-1} \\ \nu ={\left[M^0{L}^0{T}^1\right]}^{-1} \\ \nu =\left[M^0{L}^0{T}^{-1}\right] \end{gathered}$$

From equation (1), we get:

$$\begin{gathered} E=h\nu \\ h={\displaystyle \frac{E}{\nu }} \end{gathered}$$

Now, we substitute the dimensions of energy and frequency, in the above expression of Planck’s constant and we get:
 $$\begin{gathered} h={\displaystyle \frac{\left[M^1{L}^2{T}^{-2}\right]}{\left[M^0{L}^0{T}^{-1}\right]}} \\ h=\left[M^1{L}^2{T}^{-1}\right] \end{gathered}$$
Hence, the dimension of Planck’s constant is found to be $$\left[M^1{L}^2{T}^{-1}\right]$$ .
The correct option is B.

Additional information:
The dimension formula shows the specific quantities in the physical quantity given. In square brackets, it is represented. If we know the physical quantities used in the calculation, it is possible to use dimensional analysis to determine how the powers of these quantities would differ.

Note:
 It is important to remember that, if in a physical quantity, a specific quantity is not present, suppose in case of acceleration, there is no significance of mass in it, so we will raise a power by $$0$$ . In the case of dimensional formula for an angle, it will be $$\left[M^0{L}^0{T}^0\right]$$ .