Question

An electron in an unknown energy level of a hydrogen atom transitions to the n=2 level and emits a photon with wavelength 410 nm in the process. What was the initial energy level? Use R[infinity]=2.179×10−18J for the hydrogen atom Rydberg constant. Use h=6.626×10−34 Js for Planck's constant. Use c=2.998×108ms for the speed of light. Your answer should be a whole number.

Answer 1

Answer:

THE INITIAL ENERGY LEVEL OF THE ELECTRON IS 1

Explanation:

Basically, the electron from the hydrogen atom moves from an unknown energy level to energy level 2, and the formula for calculating wavelength when the energy levels are given is;

$\frac{1}y} = Rh Z^{2} (\frac{1}{n1^{2} } - \frac{1}{n2^{2} })$

y = wavelength = 410 nm = 410 *10^-9 m

Rh = Rydberg constant of hydrogen atom = 1.097 *10^7 m^-1

Z = atomic number of hydrogen = 1

n1 = initial energy level = ?

n2 = final energy level = 2

Substituting the values into the equation we get;

1  / 410 * 10^-9 = (1.097 *10^7) * (1^2) * ( 1 / n1^2 - 1 / 2^2)

1/ (410 *10^-9) (1.097 *10^7)  = 1 /n1^2 - 1/ n2^2

1 / 4.4977 = 1/ n1^2 - 1/ 4

i/n1^2 = 1/4.4977 + 1/4

1 /n1^2 = 8.4977 / 17.9908

n1^2 = 17.9908 /8.4977;

n1^2 = 2.1171

n1 = square root of 2.1171

n1 = 1.455

n1 is approximately ≅ 1

So therefore the initial energy level of the electron in the hydrogen atom is 1