Question

The longest wavelength doublet absorption transition is observed at 589 and 589.6nm. Energy difference between two excited states is :

• A. $3.31\times {10}^{-22}kJ$
• B. $3.31\times {10}^{-22}J$
• C. $2.98\times {10}^{-21}J$
• D. $3.0\times {10}^{-21}kJ$

Answer 1

Answer: (B)

$3.31\times {10}^{-22}J$

Energy diffrence between two excited state is given as:

$\Delta E=E_2-E_1$

or $\Delta E=\frac{hc}{{\lambda }_2}-\frac{hc}{{\lambda }_1}$

or $\Delta E=hc(\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1})$

where $c=3\times {10}^8$ m/s and $h=6.626\times {10}^{-34}$ Js

Given: Wavelength, ${\lambda }_1=589.6nm=589.6\times {10}^{-9}m$ and ${\lambda }_2=589nm=589\times {10}^{-9}m$

Upon substitution we get:

$\Delta E=6.626\times {10}^{-34}\times 3\times {10}^8\times (\frac{1}{589\times {10}^{-9}}-\frac{1}{589.6\times {10}^{-9}})$

$\Delta E=6.626\times {10}^{-34}\times 3\times {10}^8\times (1.728\times {10}^3)$

$\Delta E=3.43\times {10}^{-22}$ J