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JEE Main 201510 AprilOnline
Exam
JEE Main
Year
2015
Date
10 April
Mode
Online
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2015 · 10 April · Online · Physics · Units and Dimensions

If the capacitance of a nanocapacitor is measured in terms of a...

If the capacitance of a nanocapacitor is measured in terms of a unit uu, made by combining the electronic charge ee, Bohr radius a0,a0, Planck's constant hh and speed of light cc then

Options

  1. Au=e2a0hcu = \dfrac{{e}^{2}{a}_{0}}{hc}
  2. Bu=hce2a0u = \dfrac{hc}{{e}^{2}{a}_{0}}
  3. Cu=e2cha0u = \dfrac{{e}^{2}c}{h{a}_{0}}
  4. Du=e2hca0u = \dfrac{{e}^{2}h}{c{a}_{0}}

Correct answer

A. u=e2a0hcu = \dfrac{{e}^{2}{a}_{0}}{hc}

Step-by-step solution

∵ CapacitanceC = \dfrac{Q}{∆ V}<br/><br/>Also,[hcλ]=[hca0]=[Energy]<br /><br />Also, \left[\dfrac{hc}{\lambda}\right] = \left[\dfrac{hc}{{a}_{0}}\right] = \left[Energy\right].

∴[C]=[Q][∆V]=[Q][Q][∆V][Q]∴ \left[C\right] = \dfrac{\left[Q\right]}{\left[∆ V\right]} = \dfrac{\left[Q\right]\left[Q\right]}{\left[∆ V\right]\left[Q\right]}

∵W=q∆V⇒[Q][∆V]=[Energy]∵ W = q ∆ V ⇒ \left[Q\right]\left[∆ V\right] = \left[Energy\right]

∴[C]=[Q2][Energy]=[Q2][a0][hc]∴ \left[C\right] = \dfrac{\left[{Q}^{2}\right]}{\left[Energy\right]} = \dfrac{\left[{Q}^{2}\right]\left[{a}_{0}\right]}{\left[hc\right]}

∴[Capacitance]=[Q2][a0][hc]∴ \left[Capacitance\right] = \dfrac{\left[{Q}^{2}\right]\left[{a}_{0}\right]}{\left[hc\right]}

⇒u=e2a0hc⇒ u = \dfrac{{e}^{2}{a}_{0}}{hc}
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