MCQ

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

#### Options

hydrogen

deuterium

He

^{+}Li

^{++}

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#### Solution

Li^{++}

The radius of the *n*th orbit in one electron system is given by

`r_n = (n^2a_0)/Z`

Here, *a*_{0} = 53 pm

For the shortest orbit,*n* = 1

For hydrogen,*Z* = 1

∴ Radius of the first state of hydrogen atom = 53 pm

For deuterium,

Z= 1

∴ Radius of the first state of deuterium atom = 53 pm

For He^{+},*Z* = 2

∴ Radius of He^{+} atom =`53/2 pm = 26.5 "pm"`

For Li^{++},*Z* = 3

∴ Radius of Li^{++} atom = `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^{++}.

Concept: The Line Spectra of the Hydrogen Atom

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