WBJEE 2024 Physics & Chemistry Paper

To determine which quantity has the dimension of length, we need to analyze the dimensions of each option provided. We have:$-h$ as Planck's constant- $m$ as the mass of the electron$-c$ as the velocity of lightFirst, we need to recall the dimensions of these constants:- Planck's constant (h) has the dimension of action, which is energy multiplied by time:$[h]=[E][T]$where $[E]$ is the dimension of energy and $[T]$ is the dimension of time. - The dimension of energy $(E)$ is given by $[E]=\frac{[M][L]^2}{[T]^2}$, where $[M]$ is mass, $[L]$ is length, and $[T]$ is time.- The dimension of mass $(m)$ is simply $[M]$.- The dimension of the velocity of light $(c)$ is $[c]=\frac{[L]}{[T]}$.We can calculate the dimensions for each option:Option A:$\;\frac{hc}{m}$$\;\frac{hc}{m}$Substituting the known dimensions:$\left[\;\frac{hc}{m}\right] = \;\frac{\frac{[E][T][L]}{[T]}}{[M]} = \;\frac{\frac{[M][L]^2}{\frac{[T]^2[L]}{[T]}}}{[M]} = [L]^3$Option B:$\;\frac{h}{(mc)^2}$ Substituting the known dimensions:$\left[\;\frac{h}{(mc)^2}\right] = \frac{[E[T]}{\frac{[M][L]^2}{[T]^2}} = \frac{\frac{[M][L]^2}{[T]^2[T]}}{\frac{[M][L]^2}{[T]^2}} = [T] = \left[\frac{1}{c}\right]^{-1}[L]$ Option D:$\;\frac{h}{mc}$Substituting the known dimensions:$\left[\;\frac{h}{mc}\right] = \frac{[E][T]}{\frac{[M][L]}{[T]}} = \frac{\frac{[M][L]^2}{[T]^2[T]}}{\frac{[M][L]}{[T]}} = [L]$Thus, the quantity that has the dimension of length is Option D:$\;\frac{h}{mc}$