Question

If the energy difference between the electronic states is 214.68 kJ mol${}^{-1}$, calculate the frequency of light emitted when an electron drops form the higher to the lower state Plank's constant?

h = 39.79 $\times$ kJ mol${}^{-1}$

• A. 5.39 $\times$ 10${}^{14}$ cps
• B. 5.99 $\times$ 10${}^{15}$ cps
• C. 5.99 $\times$ 10${}^{14}$ cps
• D. 5.39 $\times$ 10${}^{15}$ cps

Answer 1

The correct option is A 5.39 $\times$ 10${}^{14}$ cps

$\Delta E=E_2-E_1=hv$

$\therefore 214.68=39.79\times {10}^{-14}\times v$

$v=5.39\times {10}^{14}$ cps