The energy levels of an atom are as shown in figure . Which one of those transition will result in the emission of a photon of wavelength `275 nm`?

Explanation

(a) For element A
Ground state energy, `E_(1) = -2eV`
Excited state energy, `E_(2) = 0 eV`
Energy of photon emitted, `E = E_(2) - E_(1)`
`= 0 - (-2) = 2 eV`
`:.` Wavelength of photon emitted,
`lamda = (hc)/(E) = (6.626 xx 10^(-34) xx 3xx 10^(8))/(2 xx 1.6 xx 10^(-19)) = (19.878 xx 10^(-7))/(3.2) = 6.211 xx 10^(-7) m = 621 : 1nm`
For element B
`E_(1) =4.5 eV, E_(2) = 0eV`
`E = 0 - (-4.5) = 4.5 eV`
`:. lamda = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(4.5 xx 1.6 xx 10^(-19)) = (19.878 xx 10^(-7))/(7.2) = 2.760 xx 10^(-7) = 276 nm`
For element C `E_(1) = -4.5 eV, E_(2) = -2 eV`
`E = -2- (-4.5) = -2 + 4.5 = 2.5 eV`
`:. " " lamda = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(2.5 xx 1.6 xx 10^(-19))`
`= (19.878 xx 10^(-7))/(4) = 4.969 xx 10^(-7) m = 496.9 nm`
For element D `E_(1) = -10 eV, E_(2) = -2eV`
`E = -2-(-10) = 8 eV`
`:. lamda (6.626 xx 10^(-34) xx 3 xx 10^(8))/(8xx 1.6 xx 10^(-19)) = (19.878 xx 10^(-7))/(1.28) = 1.552 xx 10^(-7) m = 155.2 nm`
`:. Element B has a photon of wavelength 275 nm
(b) Element A has radiation of maximum wavelength 621 nm