JEE PYQ: Units & Measurements - Question ID f8866c238c8c (JEE Main 2024)

The de-Broglie wavelength $\lambda$ of a particle links its wave-like behavior to its momentum $p$ via $\lambda = \frac{h}{p}.$ For a non-relativistic particle of mass $m$ and kinetic energy $E$, the momentum is given by $p = \sqrt{2\,m\,E}.$ Substituting into the de-Broglie relation yields the given expression $\lambda = \frac{h}{\sqrt{2\,m\,E}}.$ Planck’s constant $h$ is the proportionality factor between energy and angular frequency (or between momentum and wave number). Its dimensional formula must ensure that both sides of the de-Broglie relation have the same dimensions.

1. Write the given relation dimensionally: $[\lambda] = \frac{[h]}{\bigl[\,\sqrt{2\,m\,E}\,\bigr]}.$
2. Express each quantity in terms of mass $[M]$, length $[L]$, and time $[T]$:
   • Wavelength $\lambda$ has dimension $[L]$.
   • Mass $m$ has dimension $[M]$.
   • Energy $E$ has dimension $[M\,L^2\,T^{-2}]$.
3. Compute the denominator’s dimension: $\bigl[\,\sqrt{2\,m\,E}\,\bigr] = \sqrt{[M]\cdot[M\,L^2\,T^{-2}]} = \sqrt{[M^2\,L^2\,T^{-2}]} = [M\,L\,T^{-1}].$
4. Equate dimensions on both sides: $[L] = \frac{[h]}{[M\,L\,T^{-1}]}.$ Multiply both sides by $[M\,L\,T^{-1}]$ to isolate $[h]$: $[h] = [L]\cdot[M\,L\,T^{-1}] = [M\,L^2\,T^{-1}].$
5. Compare with the given options. Option C is $[M\,L^2\,T^{-1}]$, which matches our result.

Students often confuse the dimensions of energy $[M\,L^2\,T^{-2}]$ with those of Planck’s constant. A frequent mistake is to think $h$ has the same dimensions as energy or as angular momentum (which is also $[M\,L^2\,T^{-1}]$ but arises from a different context). Always derive dimensions from a known physical relation rather than memorizing them.