Question

The lines on Absorption Atomic Spectra correspond to energies needed for electrons to be excited from a lower energy level to a higher energy level. Assume that the energy needed for an electron in 2p orbital in an O atom to jump to 3s orbital is 3.6*10^-19 J, what is its wavelength of the line atomic spectra in nanometer (nm)?

Note: please use whole numbers and 3 sig figs, or no decimal place.

Answer 1

553 nm

Step-by-step explanation:

When an electron from O absorbs radiation with an energy (E) of 3.6 × 10⁻¹⁹ J, it is excited from orbital 2p to orbital 3s. The wavelength (λ) associated with that radiation can be calculated using the Planck-Einstein equation.

E = h. ν = h . c . λ⁻¹

where,

h is the Planck's constant

c is the speed of light

ν is the frequency

`\lambda = (h.c)/(E) =(6.63 * 10^(-34)J.s * 3.00 * 10^(8) m/s)/(3.6 * 10^(-19)J ) .(10^(9)nm )/(1m) =553nm`