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Question: The quantity \(\dfrac{{{e}^{2}}}{2h{{\varepsilon }_{0}}c}\) has a value \(A)\text{ }\dfrac{1}{137}...

The quantity e22hε0c\dfrac{{{e}^{2}}}{2h{{\varepsilon }_{0}}c} has a value
A) 1137ms−1A)\text{ }\dfrac{1}{137}m{{s}^{-1}}
B) 2137ms−1B)\text{ }\dfrac{2}{137}m{{s}^{-1}}
C) 1137C)\text{ }\dfrac{1}{137}
D) 2137D)\text{ }\dfrac{2}{137}

Explanation

Solution

In the expression, ee is the magnitude of the charge on a proton, hh is the planck’s constant, ε0{{\varepsilon }_{0}} is the permittivity of vacuum and cc is the speed of light in vacuum. By putting the values of all of these quantities in the expression, we can get the required value of the expression.

Complete step-by-step answer:
We will plug in the values of the physical quantities in the total expression and then find out the value of the total expression. The given expression is e22hε0c\dfrac{{{e}^{2}}}{2h{{\varepsilon }_{0}}c}
In the expression, e=1.602×10−19Ce=1.602\times {{10}^{-19}}C is the charge on a proton.
h=6.636×10−34Jsh=6.636\times {{10}^{-34}}Js is the planck’s constant.
ε0=8.854×10−12m−3kg−1s4A2{{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{m}^{-3}}k{{g}^{-1}}{{s}^{4}}{{A}^{2}} is the permittivity of free space.
c=3×108m.s−1c=3\times {{10}^{8}}m.{{s}^{-1}} is the speed of light in vacuum.
Putting these values in the quantity, we get
e22hε0c=(1.602×10−19)22×6.636×10−34×8.854×10−12×3×108=1137\dfrac{{{e}^{2}}}{2h{{\varepsilon }_{0}}c}=\dfrac{{{\left( 1.602\times {{10}^{-19}} \right)}^{2}}}{2\times 6.636\times {{10}^{-34}}\times 8.854\times {{10}^{-12}}\times 3\times {{10}^{8}}}=\dfrac{1}{137}
Therefore the required value of the expression e22hε0c\dfrac{{{e}^{2}}}{2h{{\varepsilon }_{0}}c} is 1137\dfrac{1}{137}.
Therefore, the correct option is C) 1137C)\text{ }\dfrac{1}{137}.

So, the correct answer is “Option C”.

Note: Even more than the calculation of the numerical values, students should be careful while determining the units of the total quantity. A good approach to avoiding confusion and error while calculating the final units of the expression would be to convert the units of all the physical quantities in the expression in the base fundamental SI units. This would make it much easier to calculate the combinations of the units because there are only a small finite number of fundamental units. However, many complex units of the physical quantities will make the problem harder as it will be difficult to calculate the combination of different complex units of the different physical quantities with each other.