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If the capacitance of a nanocapacitor is measured in terms of a unit $$u$$, made by combining the electronic charge $$e$$, Bohr radius $$a_0$$, Planck's constant $$h$$ and speed of light $$c$$ then
The capacitance $$ u $$ is to be expressed in terms of the electronic charge $$ e $$, Bohr radius $$ a_0 $$, Planck's constant $$ h $$, and the speed of light $$ c $$. We use dimensional analysis to find the correct expression.
First, recall the dimensions of each quantity in terms of mass (M), length (L), time (T), and current (I):
The capacitance $$ u $$ has the dimension $$[M^{-1} L^{-2} T^4 I^2]$$. We assume $$ u $$ is proportional to $$ e^a \times a_0^b \times h^c \times c^d $$, so the dimensional equation is:
$$ [M^{-1} L^{-2} T^4 I^2] = [I T]^a \times [L]^b \times [M L^2 T^{-1}]^c \times [L T^{-1}]^d $$
Expanding the right side:
$$ [I T]^a = I^a T^a $$
$$ [L]^b = L^b $$
$$ [M L^2 T^{-1}]^c = M^c L^{2c} T^{-c} $$
$$ [L T^{-1}]^d = L^d T^{-d} $$
Combining these, the right side becomes:
$$ M^c \times L^{b + 2c + d} \times T^{a - c - d} \times I^a $$
Equating dimensions with the left side:
For mass (M): $$ c = -1 $$
For current (I): $$ a = 2 $$
For time (T): $$ a - c - d = 4 $$
Substituting $$ a = 2 $$ and $$ c = -1 $$:
$$ 2 - (-1) - d = 4 $$
$$ 2 + 1 - d = 4 $$
$$ 3 - d = 4 $$
$$ -d = 1 $$
$$ d = -1 $$
For length (L): $$ b + 2c + d = -2 $$
Substituting $$ c = -1 $$ and $$ d = -1 $$:
$$ b + 2(-1) + (-1) = -2 $$
$$ b - 2 - 1 = -2 $$
$$ b - 3 = -2 $$
$$ b = 1 $$
Thus, the exponents are $$ a = 2 $$, $$ b = 1 $$, $$ c = -1 $$, $$ d = -1 $$. The expression for capacitance is:
$$ u \propto e^2 \times a_0^1 \times h^{-1} \times c^{-1} = \frac{e^2 a_0}{h c} $$
Comparing with the options:
Hence, the correct answer is Option A.
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