Chemistry Problem

Question

For a hydrogen atom, calculate the energy of the photon that would be emitted for the orbital transition of n(initial) = 4 to n(final) = 1. The Rydberg constant is 1.09678 x 10⁷ m⁻¹. Report to four significant figures.

Answer 1

1/λ = 1.09678x10⁷m^-1 (1/n_f² - 1n_i²)

λ = wavelength = ?

n_f = final energy level = 1

n_i = initial energy level = 4

1/λ = 1.09678x10⁷m^-1 (1/1 - 1/16) = 1.09678x10⁷m^-1 (0.9375)

1/λ = 1.0282x10⁷m^-1

λ = 9.7254x10^-8 m

E = hc/λ

E = energy = ?

h = Plank's constant = 6.626x10^-34 Jsec

c = speed of light = 3x10⁸ ms^-1

λ = 9.7254x10^-8 m

E = (6.626x10^-34 Jsec)(3x10⁸ msec^-1) / (9.7254x10^-8 m)

E = 2.044x10^-18 J