Question

Interactive LearningWare 30.1 reviews the concepts that play roles in this problem. A hydrogen atom emits a photon that has momentum 0.5168 × 10-27 kg·m/s. This photon is emitted because the electron in the atom falls from a higher energy level into the n = 3 level. What is the quantum number of the level from which the electron falls? Use values of h = 6.626 × 10-34 J·s, c = 2.998 × 108 m/s, and e = 1.602 × 10-19 C.

Answer 1

5

Step-by-step explanation:

λ (wavelength) = h (Plancks) / p (momentum)

also,

λ = h*c / ΔE so

h/p = (h * c) / ΔE

p * c (speed of light) = ΔE (in joules)

[ 0.5168 x 10^-27 kg m / sec ] * 2.998 x 10^8 m / sec = 1.55 J

Converting it to electron volts:

1.55 J / ( 1.602 x 10^-19 J / eV) = 0.97 eV for the photon energy

Using the Balmer Series:

ΔE = 13.6 eV [ 1 / Nf^2 - 1 / Ni^2]

0.97 / 13.6 = [ 1 / 3^2 - 1 / Ni^2 ]

0.07132 = [(1/9) - (1/Ni^2)]

0.07132 = [0.111 - (1/Ni^2)]

0.07132 - 0.111111 = - 1/Ni^2

-0.03979 = -1 / Ni^2

Ni^2 = 1/0.03979

Ni^2 = 25

Ni = 5

The use of Planck's constant is not necessary here.

N = 5 was the original energy level.