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

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We have two wavelengths of \lambda_{1} = 589\ nm= 589\times10^{-9}m and \lambda_{2} = 589.6\ nm = 589.6\times10^{-9}m.

Calculating the frequency for each:

\nu_{1} = \frac{c}{\lambda_{1}} = \frac{3.0\times10^{8}m/s}{589\times10^{-9}m} = 5.093\times10^{14}s^{-1}  

\nu_{2} = \frac{c}{\lambda_{2}} = \frac{3.0\times10^{8}m/s}{589.6\times10^{-9}m} = 5.088\times10^{14}s^{-1}

Therefore, the energy difference between two excited states will be:

\triangle E =E_{2}-E_{1} = h(\nu_{2}-\nu_{1})

= (6.626\times10^{-34}Js)(5.093-5.088)\times10^{14}s^{-1}

= 3.31\times10^{-22}J

Posted by

Divya Prakash Singh

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