JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 15)
If e is the electronic charge, c is the speed of light in free space and h is Planck's constant, the quantity $${1 \over {4\pi {\varepsilon _0}}}{{|e{|^2}} \over {hc}}$$ has dimensions of :
$$[ML{T^{ - 1}}]$$
$$[ML{T^0}]$$
$$[{M^0}{L^0}{T^0}]$$
$$[L{C^{ - 1}}]$$
Explanation
Given
e = electronic charge
c = speed of light in free space
h = Planck's constant
We know, E = $${{hc} \over \lambda }$$
and $$F = {1 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {{d^2}}}$$ $$ \Rightarrow $$ $${{{q^2}} \over {4\pi {\varepsilon _0}}} = F{d^2}$$
$${1 \over {4\pi {\varepsilon _0}}}{{{e^2}} \over {hc}} $$
= $${{F{d^2}} \over {E\lambda }}$$
= $${{Fd.d} \over {E\lambda }}$$
= $${d \over \lambda }$$
$$ = $$ dimensionless
$$ = \left[ {{M^0}{L^0}{T^0}} \right]$$
e = electronic charge
c = speed of light in free space
h = Planck's constant
We know, E = $${{hc} \over \lambda }$$
and $$F = {1 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {{d^2}}}$$ $$ \Rightarrow $$ $${{{q^2}} \over {4\pi {\varepsilon _0}}} = F{d^2}$$
$${1 \over {4\pi {\varepsilon _0}}}{{{e^2}} \over {hc}} $$
= $${{F{d^2}} \over {E\lambda }}$$
= $${{Fd.d} \over {E\lambda }}$$
= $${d \over \lambda }$$
$$ = $$ dimensionless
$$ = \left[ {{M^0}{L^0}{T^0}} \right]$$
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