True / False

The radius of the first orbit of the Hydrogen atom is same as that of $$He^+$$.

Solution

$$r=0.53A°\times \cfrac { { n }^{ 2 } }{ Z } $$
$$\therefore { r }_{ H }=0.53\times \cfrac { { (1) }^{ 2 } }{ 1 } =0.53A°$$
$${ r }_{ { He }^{ + } }=0.53\times \cfrac { { (1) }^{ 2 } }{ 2 } =0.265A°$$
Thus given statement is false.

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