Question

The energy level diagram of an element is given below. Identify, by doing necessary calculations , which transition corresponds of the emission of a spectral line of wavelength 102.7 nm.

Answer 1

`lambda=102.7 nm = 102.7xx10^(-9)m`
The energy of the emitted photons is,
`E=(hc)/(lambda)=(6.626xx10^(-34)xx3xx10^(8))/(102.7xx10^(-9))=(19.878xx10^(-26))/(102.7xx10^(-9))=1.9355xx10^(-18)J`
`:.` Energy corresponds `=(1.9355xx10^(-18))/(1.6xx10^(-19))`
`eV=12.097 ev ~ 12.1 eV`.