Question 66: Units , Dimensions and Error Analysis

Question

The de-Broglie wavelength associated with a particle of mass $$m$$ and energy $$E$$ is $$h / \sqrt{2 m E}$$. The dimensional formula for Planck's constant is :

Accepted Answer

To determine the dimensional formula for Planck's constant, we will start by analyzing the given de-Broglie wavelength equation:
$$\lambda = \frac{h}{\sqrt{2 m E}}$$
Here, $$\lambda$$ is the wavelength, $$h$$ is the Planck's constant, $$m$$ is the mass of the particle, and $$E$$ is the energy of the particle.
First, let's derive the dimensional formula for each term involved:
1. Wavelength $$\lambda$$ has the dimensional formula of length $$[L]$$.
2. Mass $$m$$ has the dimensional formula $$[	ext{M}]$$.
3. Energy $$E$$ has the dimensional formula of work, which is force times distance:
$$[E] = [F][L] = [	ext{MLT}^{-2}][L] = [	ext{ML}^{2}	ext{T}^{-2}]$$.
Now, let's rewrite the equation in terms of the dimensions:
$$[L] = \frac{[h]}{\sqrt{2 [	ext{M}] [	ext{ML}^{2}	ext{T}^{-2}]}}$$
Simplifying inside the square root:
$$[L] = \frac{[h]}{\sqrt{2 [	ext{M}] [	ext{M}] [	ext{L}^{2}	ext{T}^{-2}]}}$$
$$[L] = \frac{[h]}{\sqrt{2 [	ext{M}^{2}] [	ext{L}^{2}	ext{T}^{-2}]}}$$
Since the constants like 2 do not affect the dimensional formula, we can simplify further:
$$[L] = \frac{[h]}{[	ext{M L T}^{-1}]}$$
Cross multiplying to solve for the dimensional formula of $$h$$:
$$[h] = [L] [	ext{M L T}^{-1}]$$
$$[h] = [	ext{M L}^{2} 	ext{T}^{-1}]$$
Therefore, the dimensional formula for Planck's constant $$h$$ is:
$$\left[	ext{ML}^{2}	ext{T}^{-1}ight]$$
Hence, the correct option is:
Option C $$\left[\mathrm{ML}^2 \mathrm{~T}^{-1}ight]$$

Answer

$$\left[\mathrm{M}^2 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$$

Answer

$$\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$$

Answer

$$\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$$

Answer

$$\left[\mathrm{MLT}^{-2}\right]$$

Units , Dimensions and Error Analysis

The de-Broglie wavelength associated with a particle of mass $m and energy E is h / \sqrt{2 m E}$. The dimensional formula for Planck's constant is :

Choose the correct answer: