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

We are given the equation for the de-Broglie wavelength:

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

From this equation, rearranging to solve for \( h \):

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

Now, let's find the dimensional formula for \( h \):

• The dimensional formula for \( \lambda \) (wavelength) is \([L]\).
• The dimensional formula for \( m \) (mass) is \([M]\).
• The dimensional formula for \( E \) (energy) is \([ML^2T^{-2}]\).

Now, substitute these into the equation:

\[ [h] = [L] \times \sqrt{[M] \times [ML^2T^{-2}]}. \]

Simplifying the terms inside the square root:

\[ [h] = [L] \times \sqrt{[M] \times [M][L^2][T^{-2}]} = [L] \times \sqrt{[M^2L^2T^{-2}]} = [L] \times [MLT^{-1}]. \]

Thus, the dimensional formula for \( h \) is:

\[ [h] = [ML^2T^{-1}]. \]

Therefore, the correct answer is Option (2).