In a hydrogen like ionized atom a single electron is orbiting around the stationary positive charge. If a spectral line of `lambda` equal to `4861Å` is observed due to transition from `n=12` to `n-6`. What is the wavelength of a spectral line due to transition from `n=9` to `n=6` and also identify the element.

Answer

`6563Å, Z=3`

Explanation

Let the ground state energy for the given atom `=E_(0)`
for the transition `n_(i)=12n_(f)=6`
`DeltaE_(1)=E_(0)((1)/(n^(2))-(1)/(m^(2)))=E_(0) [((1)/(2))^(2)-((1)/(6))^(2)]`
for the transition `n_(i)=9 n_(f)=6`
`DeltaE_(2)=E_(0)((1)/(81)-(1)/(36))`
`DeltaE_(1)=(hc)/(lambda)`
`DeltaE_(1)=(hc)/(lambda_(1))=(hc)/(lambda)`
`DeltaE_(1)=(hc)/(lambda_(1))=E_(0)((1)/(144)-(1)/(36))=(-3E_(0))/(4xx36)`.....(i)
`DeltaE_(2)=(hc)/(lambda_(2))=E_(0)((1)/(81)-(1)/(36))=(-5E_(0))/(9xx36)`....(i)
Dividing (i) by (ii)
`(lambda_(2))/(lambda_(1))=(27)/(20)`
`lambda_(2)=(27)/(20)xx9861Å`
`=6563Å`
`-(3E_(0))/(4xx36)=(hc)/(4861Å)=2.55 eV`
`E_(0)-Z^(2)xx13.6eV`. (`13.6` ground state energy for hydrogen atom).
As `E_(0)` is the ground state energy for the energy given ion
`:. E_(0)= -Z^(2)xx13.6 eV. or Z^(2)=9` or
`Z=3`