Question:
If e is the electronic charged, $\mathrm{c}$ is the speed of light in free space and $\mathrm{h}$ is planck's constant, the quantity $\frac{1}{4 \pi \varepsilon_{0}} \frac{|e|^{2}}{\mathrm{hc}}$ has dimensions of:
Correct Option: , 2
Solution:
(2)
Given $e=$ electronic charge $\mathrm{c}=$ speed of light in free space $\mathrm{h}=$ planck's constant $\frac{1}{4 \pi e_{0}} \frac{e^{2}}{h c}=\frac{k e^{2}}{h c} \times \frac{\lambda^{2}}{\lambda^{2}}$
$=\frac{F \times \lambda}{E}$
$=\frac{E}{E}$
$=$ dimensionless $=\left[M^{0} L^{0} T^{0}\right]$
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