The radius of the shortest orbit in a one-electron system is 18 pm. It may be

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KollegeApply · Accepted Answer

Li++ The radius of the nth orbit in one electron system is given by r_n = (n^2a_0)/Z Here, a0 = 53 pm For the shortest orbit, n = 1 For hydrogen, Z = 1 $\therefore$ Radius of the first state of hydrogen atom = 53 pm For deuterium, Z= 1 $\therefore$ Radius of the first state of deuterium atom = 53 pm For He+, Z = 2 $\therefore$ Radius of He+ atom =$\frac{53}{2}$ pm = 26.5 "pm" For Li++, Z = 3 $\therefore$ Radius of Li++ atom = `$\frac{53}{3}$ "pm" = 17.66"pm" ≈ 18 "pm"` The given one-electron system having radius of the shortest orbit to be 18 pm may be Li++. The given one-electron system having radius of the shortest orbit to be 18 pm may be Li++.

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