A) 1/2
B) 2/137
C) 1/137
D) 1/237
Correct Answer: C
Solution :
Speed of electron in nth orbit of hydrogen atom \[v=\frac{{{e}^{2}}}{2{{\varepsilon }_{0}}n\,h}\] In ground state n = 1 Þ \[v=\frac{{{e}^{2}}}{2{{\varepsilon }_{0}}h}\] Þ \[\frac{v}{c}=\frac{{{e}^{2}}}{2{{\varepsilon }_{0}}ch}=\frac{{{(1.6\times {{10}^{-19}})}^{2}}}{2\times 8.85\times {{10}^{-12}}\times 3\times {{10}^{8}}\times 6.6\times {{10}^{-34}}}\]\[=\frac{1}{137}\].
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