The distance between the 4^{th} and 3^{rd} Bo | vedclass

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(C) The radius of the $n^{th}$ Bohr orbit is given by the formula $r_n = 0.529 	imes \frac{n^2}{Z} \ \mathring{A}$.For $He^{+}$,the atomic number $Z

Vedclass · Accepted Answer

$1.851 	imes 10^{-10} \ m$

Vedclass · Answer

(C) The radius of the $n^{th}$ Bohr orbit is given by the formula $r_n = 0.529 	imes \frac{n^2}{Z} \ \mathring{A}$.For $He^{+}$,the atomic number $Z = 2$.The radius of the $4^{th}$ orbit $(r_4)$ is $r_4 = 0.529 	imes \frac{4^2}{2} = 0.529 	imes 8 = 4.232 \ \mathring{A}$.The radius of the $3^{rd}$ orbit $(r_3)$ is $r_3 = 0.529 	imes \frac{3^2}{2} = 0.529 	imes 4.5 = 2.3805 \ \mathring{A}$.The distance between the $4^{th}$ and $3^{rd}$ orbits is $\Delta r = r_4 - r_3 = 4.232 - 2.3805 = 1.8515 \ \mathring{A}$.Since $1 \ \mathring{A} = 10^{-10} \ m$,the distance is $1.8515 	imes 10^{-10} \ m$.

The distance between the $4^{th}$ and $3^{rd}$ Bohr orbits of $He^{+}$ is $:-$

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