Question

A Li atom absorbs with a wavelength 619.6nm, and is excited from the 2s ground state to the 2p excited state. If the first ionization of Li is 520.2kJ/mol, the wavelength of the photon required to ionize the excited Li atom is closest to

A) 365.7nm

B) 619.6nm

C) 273.5nm

D) 229.9nm

E) 38.18nm

Answer 1

Given that first ionization of Li is 520.2 kJ/mol.

E = 520.2 kJ/mol = 520.2 x 10³ J/mol

We know that

E = Nhc/λ

N = Avogadro number = 6.023 x 10²³ mol^-1

   h= planck's constant = 6.626 x 10^-34 J.s

c = velocity of light = 3 x 10 ⁸ m/s

λ = wavelength

    E = Nhc/λ

   λ = Nhc/E

= (6.023 x 10²³ mol^-1) (6.626 x 10^-34 J.s) (3 x 10⁸ m/s) / (520.2 x 10³ J/mol)

= 0.230 x 10^-6 m

= 230 x 10^-9 m

= 230 nm ( 1 nm = 10^-9 m)

λ = 230 nm   

Therefore, wavelength of the photon required to ionize the excited Li atom is closest to 230 nm .

Therefore, answer is D = 229.9 nm