Question

The de-Broglie wavelength associated with a particle of mass \( m \) and energy \( E \) is \[\lambda = \frac{h}{\sqrt{2mE}}.\]The dimensional formula for Planck's constant is:

• A. \([ML^{-1}T^{-2}]\)
• B. \([ML^2T^{-1}]\)
• C. \([MLT^{-2}]\)
• D. \([M^2L^2T^{-2}]\)

Answer 1

The Correct Option is B

The objective is to derive the dimensional formula for Planck's constant \( h \) using the de-Broglie wavelength equation:

\[\lambda = \frac{h}{\sqrt{2mE}}\]

Rearranging the equation to solve for \( h \):

\(h = \lambda \times \sqrt{2mE}\)

The dimensions of each component are:

• Wavelength \(\lambda\): \([L]\)
• Mass \(m\): \([M]\)
• Energy \(E\): \([ML^2T^{-2}]\)

The dimensions of \(\sqrt{2mE}\) are:

\(\sqrt{2mE} = \sqrt{[M][ML^2T^{-2}]} = \sqrt{[M^2L^2T^{-2}]} = [MLT^{-1}]\)

Substituting these dimensions into the formula for \( h \):

\(h = [L] \times [MLT^{-1}] = [ML^2T^{-1}]\)

Thus, the dimensional formula for Planck's constant \( h \) is:

[ML²T^-1]

This result aligns with the correct answer option provided.

The dimensional analysis confirms that the correct dimensional formula for Planck's constant is \([ML^2T^{-1}]\).