Question

When an electron in an atom of element X relaxes from the third excited state to the ground state light is emitted. The light is found to have an energy of 1260 kJ/mol. What is the wavelength, in nm (to 1 decimal place, input value only), of the light?

Answer 1

Solution.

$$\lambda = 1/(R*(1/(n^2) - 1/(k^2)))$$

n = 1

k = 3

R = 1.097*10^-7

$$\lambda = 1/(1.097*(10^-7)*(1-1/9)) = 102.6 nm$$

Answer:

$$\lambda = 102.6 nm$$