An electron in hydrogen-like atom makes a transition from nth orbit or emits radiation corresponding to Lyman series. If de Broglie wavelength of electron in nth orbit is equal to the wavelength of rediation emitted , find the value of `n` . The atomic number of atomis `11`.

Explanation

`(1)/(lambda)=Z^(2)R((1)/(1^(2))-(1)/(n^(2))) " " (i)`
`2 pi r_(r)=n lambda`
`(1)/(lambda)=(n)/(2pi r_(n))=(n)/(2pi(0.53)(n^(2))/(Z)xx10^(-10))`
`=(Z)/(2pi (0.53)nxx10^(-10)) " "(ii)`
`Z^(2)R(1-(1)/(n^(2)))=(Zxx10^(10))/(2 pi (0.53)n)`
`ZR(n-(1)/(n))=(10^(10))/(2 pi(0.53))`
`(n^(2)-)/(n)=(10^(10))/(2pi (0.53)ZR)=(10^(10))/(2pi(0.53)xx11xx1.09xx10^(7))`
=25
`n^(2)-25 n-1=0`
`n=(25+-sqrt(625+4))/(2)`
`-ve` sign is not allowed
`n~=25`