Question 6: Find the dimensions of Planck’s constant ‘h’ from formula E = hf, where E is energy and f is frequency. Gravitational constant ‘G’ from the formula F=G (m_1 m_2)/r^2 where F is force m1 and m2 are masses of the objects and ‘r’ is the distance between the centers of the objects.
(a) Dimensions of Planck's constant, h
Given that E = hf ⇒ h = E/f
Where E is the energy, f is the frequency of oscillator and ‘h’ is Planck’s constant. To find the dimensions of h, therefore, we have to find the dimensions f E and f. Now, joul is the unit of energy, the dimensions of energy can be calculated as
Dimensions of energy = [J] =[N m] = [kg ms-2 m] = [kg m2 s-2]
⇒ Dimensions of energy = [M][L2][T-2] = [ML2T-2] ….. (A)
Similarly, dimensions of frequency, f = [T-1] ….. (B)
Now to find the dimensions of Planck’s constant, h, we divide the dimensions of energy on the dimensions of frequency.
Dimensions of ‘h’ = [ML2T-2]/[T-1] = [ML2T-1]
(b) Dimensions of gravitational constant, G
Newton’s law of Universal Gravitation is
Here G is the gravitational constant. We have to find the dimensions of G. Rearranging the above equation
Now a consistent equation has same dimensions on both sides. So we find the dimensions on RHS.
Since F = ma. And dimensions of mass = [M]
Acceleration is measured in ms-2, therefore, dimensions of acceleration = [LT-2]
Therefore, dimensions of F = [M][LT-2] —– (i)
Dimensions for r2 = [L][L] = [L2] —– (ii)
Dimensions of mass = [M] —– (iii)
Now substitute the dimensions from (i), (ii) and (iii) in equation (A), we get
Dimensions of G = [M][LT-2][L2]/[M2] = [ML3T-2][M-2] = [M-1L3T-2]
There is a mistake in numerical 6 part à in the last step
Yes, thanks Faizan. T is raised to power -1. Now it is corrected. Thanks once again.
Very very thankful 💞
Your are welcome.
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