Question

NIST (National Institue of Standards and Technology, USA) maintains a database of the spectra of the elements at http://www.nist.gov/pml/ data/asd_contents.cfm. The following energy levels for neutral lithium were obtained from this database:

Electronic Configuration 1s22s 1s2p 1s23s 1s23p Level, eV 0.00000 1.847818 1.847860 3.373129 3.834258 3.834258 3.878607 3.878612 4.340942 4.521648 4.521648 4.540720 4.540723 1s23d1 152451 1s24p1 15 4d

Write the ground state electronic configuration for neutral lithium and then determine the colour of light that would be emitted by an electron going from the 7 excited states given in the above table. Which transition dominated lithium in a lithium flame test?

Answer 1

The ground state of Lithium is 1s²2p¹

We will need to find the wavelength corresponding to each transition

We will use the formula,

$E=\frac{hc}{\lambda}$

Here, E denotes the energy of transition, h is the Planks constant, c is the speed of light and $\lambda$ is the wave number. All are in SI units.

But here, Energy is given in eV and we will be calculating wavelength in nanometre

1 eV= 1.60218 x 10^-19 J

Thus, E= 1.60218 x 10^-19 E_eV

Here, E is energy in Joule and E_eV is energy in eV

Similarly, 1 nm = 10^-9 m

Thus,

X6-01 = 1

Here, $\lambda_{nm}$ denotes wavelength in nanometre.

Thus, we have,

hace 1.60218 x 10-19 Ey = (6.626176 x 10-34) (299792458) 10-9Anm 1239.8 Ev=- Anm 1239.8 → Inm = E.V

We will use the above formula to calculate wavelength.We will need energy only up to 4 decimals

Configuration	EeV	Wavelength (nm) = 1239.8/EeV
1s22p1	1.8478	670.96
1s23s1	3.3731	367.56
1s23p1	3.8342	323.35
1s23d1	3.8786	319.65
1s24s1	4.3409	285.61
1s24p1	4.5216	274.19
1s24d1	4.5407	273.04

Flame test of Lithium Yields a Red flame which corresponds to wavelength of approx 670 nm. This is the transition from the first excited state to ground state i.e. from 1s²2p¹ to the ground state 1s²2s¹