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The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum is
$1 / 2$
$2 / 137$
$1 / 137$
$1 / 237$
Solution
Speed of electron in $\mathrm{n}^{\text {th }}$ orbit of hydrogen atom
$\mathrm{v}=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{nh}}$
In ground state $\mathrm{n}=1 \Rightarrow \frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{h}}$
$\Rightarrow \frac{\mathrm{v}}{\mathrm{c}}=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{ch}}=\frac{\left(1.6 \times 10^{-19}\right)^{2}}{2 \times 8.85 \times 10^{-12} \times 3 \times 10^{8} \times 6.6 \times 10^{-34}}$
$=\frac{1}{137}$
Similar Questions
Match List $- I$ (Experiment performed) with List $-II$ (Phenomena discovered/associated) and select the correct option from the options given the lists
| List $- I$ | List $- II$ |
| $(1)$ Davisson and Genner | $(i)$ Wave nature of electrons |
| $(2)$ Millikan's oil drop experiment | $(ii)$ Charge of an electron |
| $(3)$ Rutherford experiment | $(iii)$ Quantisation of energy levels |
| $(4)$ Franck-Hertz experiment | $(iv)$ Existence of nucleus |
