An atom emits a spectral line of wavelength `lambda` when an electron makes a transition between levels of energy `E_(1) and E_(2)`. Which expression correctly relates `lambda E_(1) and E_(2)` ?A. `lambda = (h c)/(E_(1) + E_(2))`B. `lambda = (2 h c)/(E_(1) + E_(2))`C. `lambda = (2 h c)/(E_(1) - E_(2))`D. `lambda = (h c)/(E_(1) - E_(2))`

An atom emits a spectral line of wavelength `lambda` when an electron

1.	An atom emits a spectral line of wavelength `lambda` when an electron makes a transition between levels of energy `E_(1) and E_(2)`. Which expression correctly relates `lambda E_(1) and E_(2)` ?A. `lambda = (h c)/(E_(1) + E_(2))`B. `lambda = (2 h c)/(E_(1) + E_(2))`C. `lambda = (2 h c)/(E_(1) - E_(2))`D. `lambda = (h c)/(E_(1) - E_(2))`
Answer» Correct Answer - D By quantum thoery of radiation, the energy change `delta E` between energy level is proportional to the frequency of electromagnetic radiation `f` and is given by. `Delta E = h f = (h c)/)lambda)` `Hence, `lambda = (h c)/(Delta E) = (h c)/(E_(1) - E_(2))`

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