Calculate the energy (J) of the photon emitted when an electron in the hydrogen atom falls from n=5 to n=3. Use 2.178 E-18 J for the constant in the Bohr Equation. Answer should be 4 sig figs , SN

Question

There is a part 2 to this questions

Calculate the wavelength (m) of the light emitted during the transition from n=5 to n=2 .

Hint, first calculate the energy as you did in the previous problem.

Again, enter your answer in scientific notation to 4 significant figures.

Answer 1

Use the formula E = -2.178x10^-18 J (1/n_f² - 1/n_i²) where nf is final n and ni is initial n

n_f = 3

n_i = 5

E = -2.178x10^-18 J (1/9 - 1//15)

E = -2.178x10^-18 (0.1111 - 0.04)

E = -2.178x10^-18 (0.0711)

E = -1.549x10^-19 J

n_f = 2

n_i = 5

E = -2.178x10^-18 J (1/4 - 1//25)

E = -2.178x10^-18 (0.25 - 0.04)

E = -2.178x10^-18 (0.21)

E = -4.574x10^-19 J

wavelength = λ

E = hc/λ

λ = hc/E = (6.626x10^-34 Jsec)(3x10⁸ m) / 4.574x10^-19 J

λ = 4.346x10^-7 m = 434.6 nm