Dimensions of `("Planck's cosntant")/("Thermal diffusivity")` are denoted by

Options

A. L
B. M
C. T
D. `theta`

Explanation

To find the dimensions of the ratio of Planck's constant to thermal diffusivity, we will follow these steps:

### Step 1: Determine the dimensions of Planck's constant (h)
Planck's constant can be derived from the equation \( E = h \nu \), where:
- \( E \) is energy
- \( \nu \) is frequency

The dimension of energy \( E \) is given by:
\[ [E] = M L^2 T^{-2} \]
The dimension of frequency \( \nu \) is:
\[ [\nu] = T^{-1} \]

Now, rearranging the equation for Planck's constant:
\[ h = \frac{E}{\nu} \]

Substituting the dimensions:
\[ [h] = \frac{M L^2 T^{-2}}{T^{-1}} = M L^2 T^{-2} \cdot T = M L^2 T^{-1} \]

### Step 2: Determine the dimensions of thermal diffusivity (α)
Thermal diffusivity is defined as:
\[ \alpha = \frac{k}{\rho c_p} \]
where \( k \) is thermal conductivity, \( \rho \) is density, and \( c_p \) is specific heat capacity.

However, the dimension of thermal diffusivity can also be expressed as:
\[ [\alpha] = L^2 T^{-1} \]

### Step 3: Calculate the dimensions of the ratio \( \frac{h}{\alpha} \)
Now we can find the dimensions of the ratio of Planck's constant to thermal diffusivity:
\[ \frac{h}{\alpha} = \frac{M L^2 T^{-1}}{L^2 T^{-1}} = M \]

### Conclusion
The dimensions of the ratio \( \frac{h}{\alpha} \) are:
\[ [M] \]

### Final Answer
The dimensions of \( \frac{\text{Planck's constant}}{\text{Thermal diffusivity}} \) are denoted by \( [M] \).

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