Question

The ratio of product of radius and time period of
electron orbiting in 2nd and 3rd orbit of hydrogen
atom
(1)4/9
(2)16/81
(3)32/243
(4)16/243
​

Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

The ratio of product of radius and time period of electron orbiting in 2nd and 3rd orbit of hydrogen atom

$\huge\underline{\underline{\bf \orange{Solution-}}}$

We know that ➝

✪$\large{\boxed{\bf \blue{r\:\:\alpha\:\:\dfrac{n^2}{Z}}}}$

n = orbit

Z = atomic number

And also ,

✪$\large{\boxed{\bf \blue{Time\: Period (T)\:\:\alpha\:\:\dfrac{n^3}{Z^2}}}}$

$\large\underline{\underline{\sf To\:Find:}}$

• Ratio of product of radius and time period

For 2nd Orbit

Radius

$\implies{\sf r_2=\dfrac{2^2}{1}}$

$\implies{\sf \green{r_2=4} }$

Time Period

$\implies{\sf T_2=\dfrac{2^3}{1^2} }$

$\implies{\sf \green{T_2=8}}$

For 3rd Orbit

Radius

$\implies{\sf r_3=\dfrac{3^2}{1}}$

$\implies{\sf \green{r_3=9} }$

Time period

$\implies{\sf T_3=\dfrac{3^3}{1^2}}$

$\implies{\sf \green{T_3=27} }$

Ratio of product of Radius and Time period ➝

$\implies{\sf \dfrac{r_2×T_2}{r_3×T_3}=\dfrac{4×8}{9×27}}$

$\implies{\bf \red{\dfrac{r_2T_2}{r_3T_3}=\dfrac{32}{243}}}$

$\huge\underline{\underline{\bf \orange{Answer-}}}$

Option (3)

The ratio of product of radius and time period of electron orbiting in 2nd and 3rd orbit of hydrogen atom is ${\bf \red{\dfrac{r_2T_2}{r_3T_3}=\dfrac{32}{243}} }$

Answer 2

Answer:

Option 3 is the answer

32/243