question_answer 69)

The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and the energy difference between two excited states.

Answer:

\[{{\lambda }_{1}}=589nm=589\times {{10}^{-9}}m\] \[{{v}_{1}}=\frac{c}{{{\lambda }_{1}}}=\frac{3\times {{10}^{8}}}{589\times {{10}^{-9}}}=5.093\times {{10}^{14}}{{s}^{-1}}\] \[{{\lambda }_{2}}=589.6nm=589.6\times {{10}^{-9}}m\] \[{{v}_{2}}=\frac{c}{{{\lambda }_{2}}}\frac{3\times {{10}^{8}}}{589.6\times {{10}^{-9}}}=5.088\times {{10}^{14}}{{s}^{-1}}\] \[\Delta E={{E}_{1}}-{{E}_{2}}\] \[=h({{v}_{1}}-{{v}_{2}})\] \[=6.626\times {{10}^{-34}}(5.093\times {{10}^{14}}-5.088\times {{10}^{14}})\] \[=3.31\times {{10}^{-22}}J\]