Question

If the capacitance of a nanocapacitor is measured in terms of a unit $'u'$ made by combining the electronic charge $V$, Bohr radius $'a_0'$, Planck's constant $'h'$ and speed of light $'c'$ then :

• A. $u = \frac{e^{2}c}{ha_{0}}$
• B. $u = \frac{e^{2}h}{ca_{0}}$
• C. $u = \frac{e^{2}a_{0}}{hc}$
• D. $u = \frac{hc}{e^{2}a_{0}}$

Answer 1

The Correct Option is C

 To solve this problem, we need to determine how to express capacitance using the given physical constants: electronic charge \(e\), Bohr radius \(a_0\), Planck's constant \(h\), and the speed of light \(c\). Let’s evaluate each option:

1. Option 1: \(u = \frac{e^{2}c}{ha_{0}}\). This option has the dimensions of charge squared times speed per energy times length, which does not correspond to capacitance.
2. Option 2: \(u = \frac{e^{2}h}{ca_{0}}\). This option has the dimensions of charge squared times action per speed times length, which is also incorrect for capacitance.
3. Option 3: \(u = \frac{e^{2}a_{0}}{hc}\). The dimensions here are charge squared times length per action times speed. This aligns with the dimensions of capacitance, considering a dimensional capacitance check:
   • Dimensional formula for capacitance: \([C] = [Q^2]/[E]\) where \(Q\) is charge and \(E\) is energy.
   • Expressing energy using other constants: \([E] = [h][c]/[a_0]\).
   • Combining these, we see that capacitance can be dimensional cubing approximated with: \([C] = [e^2 a_0]/[hc]\), matching Option 3.
4. Option 4: \(u = \frac{hc}{e^{2}a_{0}}\). This option yields the inverse dimensional check required for capacitance.

The correct unit expression for capacitance in terms of the given constants is \(u = \frac{e^{2}a_{0}}{hc}\), and this matches Option 3.

Therefore, the correct answer is:

\(u = \frac{e^{2}a_{0}}{hc}\)