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If the radius of inner most electronic orbit of a hydrogen atom is `5.3xx10^(-11)` m, then the radii of `n=2` orbit isA. `1.12 Å`B. `2.12 Å`C. `3.22 Å`D. `4.54 Å` |
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Answer» Correct Answer - B As,`r_n=n^2a_0` Here,`a_0=5.3xx10^(-11)m`,`n=2` `therefore r_2=(2)^2a_0=4a_0=4xx5.3xx10^(-11)m` `=21.2xx10^(-11)m=2.12Å`. |
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