Question

show that the ratio of radius of the third Orbit to that of second orbit is 2.25

Answer 1

$\underline{ \boxed{ \bold{ \mathfrak{ \purple{ \huge{Answer}}}}}} \\ \\ \red{ \star} \: \mathfrak{formula \: of \: radius \: of \: orbit \: as \: per \: bohrs} \\ \mathfrak{theory \: is \: given \: as...} \\ \\ \underline{ \boxed{ \boxed{ \bold{ \rm{ \pink{r = 0.53 \times \frac{ {n}^{2} }{z} \: a \degree}}}} : \implies \boxed{ \bold{ \blue{ \rm{r \propto{ {n}^{2} }}}}} }} \\ \\ \star \: \red{ \mathfrak{calculation}} \\ \\ \leadsto \rm \: \frac{r{3}}{r{2}} = \frac{ {3}^{2} }{ {2}^{2} } = \frac{9}{4} \\ \\ \therefore \: \underline{ \boxed{ \bold{ \mathfrak{ \orange{ratio = 2.25 : 1}}}}} \: \purple{ \star}$

Answer 2

$\huge \fcolorbox{red}{pink}{Solution :)}$

We know that , the radius of nth orbit is given by

$\sf \fbox{Radius = 0.529 \times \frac{ {(n)}^{2} }{z} } \: \:$

Where ,

n = number of the orbit

z = atomic number of element

Thus , the ratio of radius of the third orbit to that of second orbit is

$\sf \hookrightarrow \frac{0.529 \times \frac{ {(3)}^{2} }{z}}{0.529 \times \frac{ {(2)}^{2} }{z}} = \frac{ {(3)}^{2} }{ {(2)}^{2} } = 2.25$

Hence , the ratio of radius of the third orbit to that of second orbit is 2.25

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