Question

Resistance has the same dimensions as, where h is the Planck's constant and e is the charge.

• A. ${\mathrm{h}}^2/{\mathrm{e}}^2$
• B. ${\mathrm{h}}^2/\mathrm{e}$
• C. $\mathrm{h}/{\mathrm{e}}^2$
• D. $\mathrm{h}/\mathrm{e}$

Answer 1

The correct option is C

$\mathrm{h}/{\mathrm{e}}^2$

Explanation for the correct option

In the case of option C,

1. According to Ohm's law, resistance can be calculated as $\mathrm{R}=\raisebox{1ex}{$\mathrm{V}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{I}$}\right.$
2. The Volt has dimension of $\left[{\mathrm{ML}}^2{\mathrm{T}}^{-3}{\mathrm{A}}^{-1}\right]$.
3. The resistance dimension is $\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-3}{\mathrm{A}}^{-1}\right]}{\left[\mathrm{A}\right]}=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-3}{\mathrm{A}}^{-2}\right]$.
4. Whereas, the dimension of $\left[\frac{\mathrm{h}}{{\mathrm{e}}^2}\right]=\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right]}{\left[{\mathrm{A}}^2{\mathrm{T}}^2\right]}=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-3}{\mathrm{A}}^{-2}\right]$
5. Thus, resistance and $\mathrm{h}/{\mathrm{e}}^2$ has same dimensional formula.

Explanation for the incorrect options

In the case of option A,

1. The dimension of ${\mathrm{h}}^2/{\mathrm{e}}^2$ is $\frac{\left[{\mathrm{M}}^2{\mathrm{L}}^4{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{A}}^2{\mathrm{T}}^2\right]}=\left[{\mathrm{M}}^2{\mathrm{L}}^4{\mathrm{T}}^{-4}{\mathrm{A}}^{-2}\right]$

In the case of option B,

1. The dimension of ${\mathrm{h}}^2/\mathrm{e}$ is $\frac{\left[{\mathrm{M}}^2{\mathrm{L}}^4{\mathrm{T}}^{-2}\right]}{\left[\mathrm{AT}\right]}=\left[{\mathrm{M}}^2{\mathrm{L}}^4{\mathrm{T}}^{-3}{\mathrm{A}}^{-1}\right]$

In the case of option D,

1. The dimension of $\mathrm{h}/\mathrm{e}$ is $\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right]}{\left[\mathrm{AT}\right]}=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}{\mathrm{A}}^{-1}\right]$.

Therefore, option C is the correct option.