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

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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