Question

The energy needed to ionize an atom of element X when it is in its most stable state is 500 kJ mol21. However, if an atom of X is in its lowest excited state, only 120 kJ mol21 is needed to ionize it. What is the wavelength of the radiation emitted when an atom of X undergoes a transition from the lowest excited state to the ground state

Answer 1

λ = 314.8 nm

Step-by-step explanation:

E = hc/λ

E: energy

h: Planck constant = 6.62607015 × 10⁻³⁴

c: speed of light = 2.998 x 10⁸ m s⁻¹

λ: wavelength

Ground state:

E(gs) = 500 000 J

Lowest excited state:

E (les) = 120 000 J

Transition energy = E(gs) - E (les)

Transition energy = 380 000 J/mol

Transition energy = 380000 J/6.022x10²³ atoms

Transition energy = 6.31 x 10⁻¹⁹ J/atom

E = hc/λ

λ = hc/E

λ = (6.62607015 × 10⁻³⁴ x 2.998 x 10⁸ m s⁻¹) / 6.31 x 10⁻¹⁹ J

λ = 3.148 x 10⁻⁷ m

λ = 314.8 nm