Units & Dimensions

JEE Main level, single correct, medium

133 of 149

Question 133 of 149

If the capacitance of a nanocapacitor is measured in terms of a unit uu, made by combining the electronic charge ee, Bohr radius a0,{a}_{0}^{}, Planck's constant hh and speed of light cc then

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Answer: (A) u=e2a0hcu=\dfrac{{e}^{2}{a}_{0}}{hc}

∵CapacitanceC=Q∆V∵CapacitanceC=\dfrac{Q}{∆V}

Also,[hcλ]=[hca0]=[Energy]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\Rightarrow \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\Rightarrow u=\dfrac{{e}^{2}{a}_{0}}{hc}

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