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 `lambda` is be Broglie wavelength , then for nth stationary orbit `2 pi r_(n) = n lambda`
where `r_(n)` is redius of nth orbit `r_(n) = (epsilon_(0) h^(2) n^(2))/(pi m Ze^(2))`
`:. 2 pi = ((epsilon_(0) h^(2) n^(2))/(pi m Ze^(2))) = n lambda, implies (1)/(lambda) = (m Ze^(2))/(2epsilon_(0) h^(2) n)` (i)
For Lyman series of hydrogen like atom
`(1)/(lambda) = Z^(2) R ((1)/(1^(2)) - (1)/(n^(2)))`
From Eqs. (i) and (ii), `Z^(2) R (1 - (1)/(n^(2))) = (m Ze^(2))/(2epsilon_(0) h^(2) n)`
Rydberg constant,
`R = (me^(4))/(8 epsilon_(0)^(2) ch^(3))`
`:. (Z^(2)me^(4))/(8 epsilon_(0)^(2) ch^(3)) = (1 - (1)/(n^(2))) = (m Ze^(2))/(2epsilon_(0) h^(2) n)`
`(1 - (1)/(n^(2))) = (4 epsilon_(0) ch)/(ne^(2) Z)`
`= (4 xx (8.85 xx 10^(-=12)) xx (3 xx 10^(8)) xx (6.62 xx 10^(-34)))/(n xx(1.6 xx 10^(-19)) xx 11) = (25)/(n)`
`:. n^(2) - 1 = 25n or n^(2) - 25n- 1 = 0`
`n = (25 +- sqrt((-25)^(2) + 4 xx 1 xx 1))/(2) = (25 +- sqrt(625))/(2) = 25`
As negative `n` is not possible , `n = 25`.