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

If `lamda` is the de Broglie wavelength then for nth orbit
`2pir_(n)=nlamda`
where `r_(n)=(epsilon_(0)h^(2)n^(2))/(pi m e^(2)Z)`
`:.1/(lamda)=(me^(2) Z)/(2 epsilon_(0)h^(2)n)`…….i
For Lyman series
`1/(lamda)=Z^(2)R(1/(1^(2))-1/(n^(2)))` ...........ii
From equations (i) and (ii) we have
`Z^(2)R(1-1/(n^(2)))=(me^(2))/(2epsilon_(0)h^(2))Z/n`..........iii
Where `R=(me^(4))/(8 epsilon_(0)^(2)ch^(3))` ..........iv
After substituting the values in iii and iv we get
`n=25`