Question

The energy difference between two electronic states is 46.12 kcal/mole.what will be the frequency the light emitted when an eletron drops from the higher to the lower energy state

Answer 1

The energy difference between two electronic states is 46.12 Kcal mol⁻¹.  

$\Delta E=h\nu=h\frac{c}{\Lambda }$  

$\Delta E$, is the energy difference  

h is the planck constant  

c is the speed of light  

${\Lambda }$ is the wavelength  

so putting in the equation  

$\Delta E$ is 46.12 Kcal mol⁻¹=192.96608 J mol⁻¹  

C is 3×10⁸ ms⁻¹  

h is 6.6 ×10 ⁻³⁴ J s

putting in the equation:  

$192.9=6.6 \times 10^{-34}\frac{3\times10^{8}}{\Lambda }$

${\Lambda } =1.02\times e^{-27}$  

$\nu=\frac{3\times 10^{8}}{1.02\times e^{-27}}$

$\nu=\frac{3\times 10^{8}}{1.02\times e^{-27}}$

$\nu=2.9\times e^{35}}$