If ratio of area enclosed in `6^(th)` orbit in H atom to 2nd orbit in `Li^(+2)` ion is `3^(x)` then find the value of x ?

Answer

1

Explanation

`2C(s)+O_(2)(g)iff2CO(g),K_(2)`
`CO_(2)(g)iffC(s)+O_(2)(g),1//K_(1)`
`C(s)+CO_(2)(g)iff2CO(g),K_(3)`
`K_(3)=(K_(2))/(K_(1))=(10^(-3))/(10^(-2))=10^(-1)`
`C(s)+O_(2)(g)iffCO_(2)(g), K_(1)`
`2CO(g)iffC(s)+CO_(2)(g)" "1//K_(3)`
`2CO(g)+O_(2)(g)iff2CO_(2)(g), K_(4)`
`K_(4)=(K_(1))/(K_(3))=(10^(-2))/(10^(-1))=10^(-1)`
`(x)/(y)=1`