Question:

Which of the following quantity has the dimension of length? (where \(h\) is Planck’s constant, \(m\) is the mass of electron and \(c\) is the velocity of light):

Show Hint

In dimensional analysis, break down the dimensions of each term in the equation and simplify them to check for consistency with the required dimension.
Updated On: Feb 15, 2025
  • \(\frac{hc}{m}\)
  • \(\frac{h}{mc^2}\)
  • \(\frac{h^2}{m^2c^2}\)
  • \(\frac{h}{mc}\)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

Step 1: First, recall the dimensional formulas for each of the physical quantities:

  • \( h \) (Planck’s constant) has the dimension \([h] = [ML^2T^{-1}]\).
  • \( m \) (mass of the electron) has the dimension \([m] = [M]\).
  • \( c \) (velocity of light) has the dimension \([c] = [LT^{-1}]\).

Step 2: Now, calculate the dimensions of each given option.

For option \( \frac{hc}{m} \):

\[ \left[ \frac{hc}{m} \right] = \frac{[ML^2T^{-1}][LT^{-1}]}{[M]} = [L^2T^{-2}] \]

which does not correspond to the dimension of length.

For option \( \frac{h}{mc^2} \):

\[ \left[ \frac{h}{mc^2} \right] = \frac{[ML^2T^{-1}]}{[M][L^2T^{-2}]} = [T] \]

which corresponds to the dimension of time, not length.

For option \( \frac{h^2}{m^2c^2} \):

\[ \left[ \frac{h^2}{m^2c^2} \right] = \frac{[M^2L^4T^{-2}]}{[M^2][L^2T^{-2}]} = [L^2] \]

which corresponds to the dimension of area, not length.

For option \( \frac{h}{mc} \):

\[ \left[ \frac{h}{mc} \right] = \frac{[ML^2T^{-1}]}{[M][LT^{-1}]} = [L] \]

which corresponds to the dimension of length.

Was this answer helpful?
0
0