In the first excited state of a hydrogen atom | vedclass

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(B) For a hydrogen atom,the radius of the $n^{th}$ orbit is given by the formula $r_n = a_0 	imes n^2$,where $a_0 = 0.529 	ext{ Å}$ is the Bohr radi

Vedclass · Accepted Answer

$2.1 	imes 10^{-10} 	ext{ m}$

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(B) For a hydrogen atom,the radius of the $n^{th}$ orbit is given by the formula $r_n = a_0 	imes n^2$,where $a_0 = 0.529 	ext{ Å}$ is the Bohr radius.For the first excited state,the principal quantum number is $n = 2$.Substituting the value of $n$ into the formula,we get $r_2 = 0.529 	imes (2)^2 	ext{ Å}$.$r_2 = 0.529 	imes 4 	ext{ Å} = 2.116 	ext{ Å}$.Since $1 	ext{ Å} = 10^{-10} 	ext{ m}$,we have $r_2 = 2.116 	imes 10^{-10} 	ext{ m}$.Rounding to two significant figures,the radial distance is approximately $2.1 	imes 10^{-10} 	ext{ m}$.

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