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):

• A. \(\frac{hc}{m}\)
• B. \(\frac{h}{mc^2}\)
• C. \(\frac{h^2}{m^2c^2}\)
• D. \(\frac{h}{mc}\)

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.

Answer 1

The Correct Option is D

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.

Answer 2

Dimensional Analysis to Find Quantity with Dimension of Length

We are asked to find which of the following quantities has the dimension of length [L], where h is Planck's constant, m is the mass of the electron, and c is the velocity of light.

The dimensions of the fundamental quantities are given as:

Planck's constant (h): $[h] = [ML^2T^{-1}]$

Velocity of light (c): $[c] = [LT^{-1}]$

Mass of electron (m): $[m] = [M]$

Checking the dimensions of each option:

(A) $\frac{hc}{m}$

Dimensions: $\frac{[ML^2T^{-1}] [LT^{-1}]}{[M]} = \frac{ML^3T^{-2}}{M} = [L^3T^{-2}]$

This does not have the dimension of length.

(B) $\frac{h}{mc^2}$

Dimensions: $\frac{[ML^2T^{-1}]}{[M] [LT^{-1}]^2} = \frac{ML^2T^{-1}}{M L^2T^{-2}} = [T^{1}]$

This has the dimension of time.

(C) $\frac{h^2}{mc^2}$

Dimensions: $\frac{[ML^2T^{-1}]^2}{[M] [LT^{-1}]^2} = \frac{M^2L^4T^{-2}}{M L^2T^{-2}} = [ML^2]$

This does not have the dimension of length.

(D) $\frac{h}{mc}$

Dimensions: $\frac{[ML^2T^{-1}]}{[M] [LT^{-1}]} = \frac{ML^2T^{-1}}{MLT^{-1}} = [L]$

This has the dimension of length.

Using the method of proportionality:

Assume Length $[L] \propto h^x c^y m^z$

$[L^1M^0T^0] = [ML^2T^{-1}]^x [LT^{-1}]^y [M]^z$

$[L^1M^0T^0] = [M^{x+z} L^{2x+y} T^{-x-y}]$

Equating the powers:

For M: $x + z = 0 \implies z = -x$

For L: $2x + y = 1$

For T: $-x - y = 0 \implies y = -x$

Substituting $y = -x$ into $2x + y = 1$:

$2x - x = 1 \implies x = 1$

Then, $y = -1$ and $z = -1$

So, $L \propto h^1 c^{-1} m^{-1} = \frac{h}{mc}$

Final Answer: (D) $\frac{h}{mc}$