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JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement

If the capacitance of a nanocapacitor is measured in terms of a unit \(u\) made by combining the electric charge \(e,\) Bohr radius \(a_0,\) Planck's constant \(h\) and speed of light \(c\) then

  1. A \(u\, = \,\frac{{{e^2}h}}{{{a_0}}}\)
  2. B \(u\, = \,\frac{{hc}}{{{e^2}{a_0}}}\)
  3. C \(u\, = \,\frac{{{e^2}c}}{{h{a_0}}}\)
  4. D \(u\, = \,\frac{{{e^2}{a_0}}}{{hc}}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(u\, = \,\frac{{{e^2}{a_0}}}{{hc}}\)

Step-by-step Solution

Detailed explanation

The unit of capacitance is \(\frac {Coulomb}{V} = \frac {Coulomb}{\text{work done per unit charge}}\)=\(\frac{\text { Coulomb }}{\frac{ W }{\text { Coulomb }}}=\frac{\text { Coulomb }^{2}}{ kgm ^{2} s ^{-2}}= C ^{2} kg ^{-1} m ^{-2} s ^{2}\) Checking units of each option:…
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